1. Key Concepts and Formulas

• Parametric Form of a Line: A point on a line passing through $(x_0, y_0, z_0)$ with direction ratios $(a, b, c)$ can be represented as $(x_0 + a\lambda, y_0 + b\lambda, z_0 + c\lambda)$ for some parameter $\lambda$.
• Collinearity of Three Points: Three points $P(x_1,y_1,z_1)$, $A(x_2,y_2,z_2)$, and $B(x_3,y_3,z_3)$ are collinear if the direction ratios of the line segment $PA$ are proportional to the direction ratios of the line segment $PB$. Mathematically, this means $\frac{x_2-x_1}{x_3-x_1} = \frac{y_2-y_1}{y_3-y_1} = \frac{z_2-z_1}{z_3-z_1}$, provided the denominators are non-zero. If any denominator is zero, the corresponding numerator must also be zero.
• Equation of a Line Passing Through Two Points: The equation of a line passing through two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.

2. Step-by-Step Solution

Step 1: Represent General Points on the Given Lines

We are given two lines: $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ $L_2: \frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}$ And a point $P(1,1,1)$.

Let $A$ be a general point on line $L_1$. We can write its coordinates in parametric form by setting the ratios equal to a parameter, say $\lambda$: $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} = \lambda$ So, $A = (1+2\lambda, -1+3\lambda, 1+4\lambda)$.

Similarly, let $B$ be a general point on line $L_2$. We set the ratios equal to another parameter, say $\mu$: $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1} = \mu$ So, $B = (3+\mu, 4+2\mu, \mu)$.

Reasoning: The line $L$ passes through $P(1,1,1)$ and intersects $L_1$ and $L_2$. This means that there exist specific points $A$ on $L_1$ and $B$ on $L_2$ such that $P$, $A$, and $B$ are collinear. Our goal is to find these specific points $A$ and $B$.

Step 2: Apply the Collinearity Condition for P, A, and B

For $P(1,1,1)$, $A(1+2\lambda, -1+3\lambda, 1+4\lambda)$, and $B(3+\mu, 4+2\mu, \mu)$ to be collinear, the direction ratios of $PA$ must be proportional to the direction ratios of $PB$.

Direction ratios of $PA$: $(x_A-x_P, y_A-y_P, z_A-z_P) = ( (1+2\lambda)-1, (-1+3\lambda)-1, (1+4\lambda)-1 )$ $= (2\lambda, 3\lambda-2, 4\lambda)$

Direction ratios of $PB$: $(x_B-x_P, y_B-y_P, z_B-z_P) = ( (3+\mu)-1, (4+2\mu)-1, \mu-1 )$ $= (2+\mu, 3+2\mu, \mu-1)$

For collinearity, these direction ratios must be proportional: $\frac{2\lambda}{2+\mu} = \frac{3\lambda-2}{3+2\mu} = \frac{4\lambda}{\mu-1}$

Reasoning: This step sets up a system of equations to solve for the unknown parameters $\lambda$ and $\mu$. We need to find values of $\lambda$ and $\mu$ that satisfy these proportionality conditions, as these values will define the specific points $A$ and $B$ that lie on line $L$.

Step 3: Solve the System of Equations for $\lambda$ and $\mu$

We can form two equations from the proportionality:

Equation 1 (from the first two ratios): $2\lambda(3+2\mu) = (3\lambda-2)(2+\mu)$ $6\lambda + 4\lambda\mu = 6\lambda + 3\lambda\mu - 4 - 2\mu$ $4\lambda\mu = 3\lambda\mu - 4 - 2\mu$ $\lambda\mu = -4 - 2\mu \quad (*)$

Equation 2 (from the last two ratios): $(3\lambda-2)(\mu-1) = 4\lambda(3+2\mu)$ $3\lambda\mu - 3\lambda - 2\mu + 2 = 12\lambda + 8\lambda\mu$ $-5\lambda\mu - 15\lambda - 2\mu + 2 = 0 \quad (**)$

Now, substitute $\lambda\mu = -4-2\mu$ from $(*)$ into $(**)$: $-5(-4-2\mu) - 15\lambda - 2\mu + 2 = 0$ $20 + 10\mu - 15\lambda - 2\mu + 2 = 0$ $8\mu - 15\lambda + 22 = 0 \quad (***)$

From $(*)$, we can express $\lambda$ in terms of $\mu$ (assuming $\mu \neq 0$; if $\mu=0$, then $0=-4$, which is false, so $\mu \neq 0$): $\lambda = \frac{-4-2\mu}{\mu}$

Substitute this expression for $\lambda$ into $(***)$: $8\mu - 15\left(\frac{-4-2\mu}{\mu}\right) + 22 = 0$ Multiply the entire equation by $\mu$ to eliminate the denominator: $8\mu^2 - 15(-4-2\mu) + 22\mu = 0$ $8\mu^2 + 60 + 30\mu + 22\mu = 0$ $8\mu^2 + 52\mu + 60 = 0$ Divide by 4: $2\mu^2 + 13\mu + 15 = 0$

Solve this quadratic equation for $\mu$ using the quadratic formula $\mu = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $\mu = \frac{-13 \pm \sqrt{13^2 - 4(2)(15)}}{2(2)}$ $\mu = \frac{-13 \pm \sqrt{169 - 120}}{4}$ $\mu = \frac{-13 \pm \sqrt{49}}{4}$ $\mu = \frac{-13 \pm 7}{4}$

Two possible values for $\mu$: $\mu_1 = \frac{-13+7}{4} = \frac{-6}{4} = -\frac{3}{2}$ $\mu_2 = \frac{-13-7}{4} = \frac{-20}{4} = -5$

Now, find the corresponding $\lambda$ values using $\lambda = \frac{-4-2\mu}{\mu}$: For $\mu = -\frac{3}{2}$: $\lambda = \frac{-4-2(-\frac{3}{2})}{-\frac{3}{2}} = \frac{-4+3}{-\frac{3}{2}} = \frac{-1}{-\frac{3}{2}} = \frac{2}{3}$

For $\mu = -5$: $\lambda = \frac{-4-2(-5)}{-5} = \frac{-4+10}{-5} = \frac{6}{-5} = -\frac{6}{5}$

We must verify which pair $(\lambda, \mu)$ satisfies the original proportionality condition. Consider the condition $\frac{2\lambda}{2+\mu} = \frac{4\lambda}{\mu-1}$. If $\lambda \neq 0$, then $\frac{1}{2+\mu} = \frac{2}{\mu-1}$, which implies $\mu-1 = 2(2+\mu) \Rightarrow \mu-1 = 4+2\mu \Rightarrow -\mu = 5 \Rightarrow \mu = -5$. If $\lambda = 0$, then $A=(1,-1,1)$. The direction ratios of $PA$ would be $(0, -2, 0)$. For $P,A,B$ to be collinear, the direction ratios of $PB$ must be proportional to $(0,-2,0)$, which means $x_B-x_P=0$ and $z_B-z_P=0$. So $2+\mu=0 \Rightarrow \mu=-2$ and $\mu-1=0 \Rightarrow \mu=1$. This is a contradiction, so $\lambda \neq 0$. Thus, the correct value for $\mu$ is $-5$, which implies $\lambda = -\frac{6}{5}$.

Let's check this pair with all three ratios: $\frac{2\lambda}{2+\mu} = \frac{2(-\frac{6}{5})}{2+(-5)} = \frac{-\frac{12}{5}}{-3} = \frac{12}{15} = \frac{4}{5}$ $\frac{3\lambda-2}{3+2\mu} = \frac{3(-\frac{6}{5})-2}{3+2(-5)} = \frac{-\frac{18}{5}-2}{3-10} = \frac{-\frac{28}{5}}{-7} = \frac{28}{35} = \frac{4}{5}$ $\frac{4\lambda}{\mu-1} = \frac{4(-\frac{6}{5})}{-5-1} = \frac{-\frac{24}{5}}{-6} = \frac{24}{30} = \frac{4}{5}$ All ratios are equal, so the values $\lambda = -\frac{6}{5}$ and $\mu = -5$ are correct.

Reasoning: Solving the system of equations for $\lambda$ and $\mu$ is the most critical algebraic step. The values obtained for $\lambda$ and $\mu$ uniquely define the points $A$ and $B$ through which line $L$ passes. It's crucial to verify the solution with all parts of the collinearity condition.

Step 4: Find the Equation of Line L

We can use the values of $\lambda$ or $\mu$ to find the coordinates of point $A$ or $B$. Let's use $\lambda = -\frac{6}{5}$ to find point $A$: $A = (1+2\lambda, -1+3\lambda, 1+4\lambda)$ $A = (1+2(-\frac{6}{5}), -1+3(-\frac{6}{5}), 1+4(-\frac{6}{5}))$ $A = (1-\frac{12}{5}, -1-\frac{18}{5}, 1-\frac{24}{5})$ $A = (\frac{5-12}{5}, \frac{-5-18}{5}, \frac{5-24}{5})$ $A = (-\frac{7}{5}, -\frac{23}{5}, -\frac{19}{5})$

Now, we have two points on line $L$: $P(1,1,1)$ and $A(-\frac{7}{5}, -\frac{23}{5}, -\frac{19}{5})$. The direction ratios of line $L$ are $(x_A-x_P, y_A-y_P, z_A-z_P)$: $DR_L = (-\frac{7}{5}-1, -\frac{23}{5}-1, -\frac{19}{5}-1)$ $DR_L = (-\frac{12}{5}, -\frac{28}{5}, -\frac{24}{5})$ We can simplify these direction ratios by multiplying by $-\frac{5}{4}$: Simplified $DR_L = (3, 7, 6)$.

The equation of line $L$ passing through $P(1,1,1)$ with direction ratios $(3,7,6)$ is: $\frac{x-1}{3} = \frac{y-1}{7} = \frac{z-1}{6} = k$ A general point on line $L$ can be written as $(1+3k, 1+7k, 1+6k)$.

Reasoning: Once the specific points $A$ and $B$ (or just one of them) are found, the problem reduces to finding the equation of a line passing through two known points ($P$ and $A$ in this case). Simplifying the direction ratios makes it easier to work with the equation and check the options.

Step 5: Check Which Option Lies on Line L

We need to check which of the given options satisfies the equation of line $L$: $\frac{x-1}{3} = \frac{y-1}{7} = \frac{z-1}{6}$.

(A) $(7,15,13)$ $\frac{7-1}{3} = \frac{6}{3} = 2$ $\frac{15-1}{7} = \frac{14}{7} = 2$ $\frac{13-1}{6} = \frac{12}{6} = 2$ Since all ratios are equal to 2, the point $(7,15,13)$ lies on line $L$.

(B) $(4,22,7)$ $\frac{4-1}{3} = 1$ $\frac{22-1}{7} = 3$ The ratios are not equal, so this point does not lie on $L$.

(C) $(10,-29,-50)$ $\frac{10-1}{3} = 3$ $\frac{-29-1}{7} = \frac{-30}{7}$ The ratios are not equal, so this point does not lie on $L$.

(D) $(5,4,3)$ $\frac{5-1}{3} = \frac{4}{3}$ $\frac{4-1}{7} = \frac{3}{7}$ The ratios are not equal, so this point does not lie on $L$.

Reasoning: The final step involves substituting the coordinates of each option into the equation of line $L$. If all three components yield the same value for the parameter $k$, then the point lies on the line.

3. Common Mistakes & Tips

• Algebraic Precision: The process of solving for $\lambda$ and $\mu$ involves multiple algebraic manipulations. Errors in signs, multiplication, or simplification can lead to incorrect values. Double-check your calculations.
• Verification of Parameters: Always substitute the calculated values of $\lambda$ and $\mu$ back into the original proportionality equations (all three parts) to ensure consistency. This helps catch errors before proceeding to find the line equation.
• Understanding Collinearity: Remember that collinearity means the direction vectors (or direction ratios) formed by any two pairs of points are parallel, hence proportional.
• Simplifying Direction Ratios: While not strictly necessary, simplifying direction ratios by dividing by a common factor can make subsequent calculations (like checking options) much easier and reduce the chance of arithmetic errors.

4. Summary

This problem required finding a point on a line $L$ that passes through a given point $P$ and intersects two other lines, $L_1$ and $L_2$. The fundamental approach was to leverage the collinearity of $P$ with a point $A$ on $L_1$ and a point $B$ on $L_2$. We first expressed general points $A$ and $B$ in their parametric forms. Then, by equating the ratios of the direction numbers of $PA$ and $PB$, we formed a system of equations in two variables, $\lambda$ and $\mu$. Solving this system yielded unique values for $\lambda$ and $\mu$, which in turn allowed us to determine the coordinates of point $A$. With point $P$ and point $A$ known, we found the direction ratios and the Cartesian equation of line $L$. Finally, we tested each given option by substituting its coordinates into the equation of line $L$ to identify the point that lies on it.

5. Final Answer

The final answer is $\boxed{(7,15,13)}$, which corresponds to option (A).