Key Concepts and Formulas

• Equation of a Line in 3D: A line passing through a point $(x_0, y_0, z_0)$ with direction ratios $(a,b,c)$ can be represented in symmetric form as $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$ or in parametric form as $x=x_0+a\lambda, y=y_0+b\lambda, z=z_0+c\lambda$.
• Distance of a Point from a Line Along Another Line: To find the distance of a point $A$ from a line $L_1$ along a line $L_2$, we need to find a point $B$ on $L_1$ such that the line segment $AB$ is parallel to $L_2$. This involves constructing a line through $A$ parallel to $L_2$, finding its intersection with $L_1$, and then calculating the distance between $A$ and this intersection point.
• 3D Distance Formula: The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Step-by-Step Solution

Step 1: Identify the Given Point and Lines We are given the point $A = (7, -2, 11)$.

The first line, $L_1$, is: $L_1: \frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ From this equation, we can identify that $L_1$ passes through the point $P_1(6,4,8)$ and has a direction vector $\vec{d_1} = (1,0,3)$.

• Reasoning: In the symmetric form $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$, $(x_0,y_0,z_0)$ is a point on the line and $(a,b,c)$ are the direction ratios. The denominator $0$ for the $y$-term implies that the $y$-coordinate for any point on $L_1$ must be $4$ (i.e., $y-4=0$).

The second line, $L_2$, is: $L_2: \frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$ From this equation, we identify that $L_2$ passes through the point $P_2(5,1,5)$ and has a direction vector $\vec{d_2} = (2,-3,6)$.

Step 2: Formulate the Equation of the Line Passing Through $A$ and Parallel to $L_2$ Let this new line be $L_A$. We are looking for the distance "along the line $L_2$", which means the line segment connecting $A$ to the target point on $L_1$ must be parallel to $L_2$. Since $L_A$ passes through $A(7,-2,11)$ and is parallel to $L_2$, its direction vector will be the same as $\vec{d_2}$, which is $(2,-3,6)$.

The parametric equation of $L_A$ can be written as: $L_A: \quad x = 7 + 2\lambda$ $y = -2 - 3\lambda$ $z = 11 + 6\lambda$ where $\lambda$ is a scalar parameter.

• Reasoning: Any point on $L_A$ can be expressed in terms of $\lambda$. We will use this to find the point where $L_A$ intersects $L_1$.

Step 3: Find the Point of Intersection $B$ of $L_A$ and $L_1$ The point $B$ lies on both $L_A$ and $L_1$. Therefore, its coordinates must satisfy the equations of both lines. We substitute the parametric coordinates of a general point on $L_A$ into the symmetric equation of $L_1$: $\frac{(7+2\lambda)-6}{1} = \frac{(-2-3\lambda)-4}{0} = \frac{(11+6\lambda)-8}{3}$

Simplify the numerators: $\frac{1+2\lambda}{1} = \frac{-6-3\lambda}{0} = \frac{3+6\lambda}{3}$

For a point to lie on $L_1$, the numerator corresponding to the zero denominator must itself be zero. Thus, from the middle term: $-6-3\lambda = 0$ $-3\lambda = 6$ $\lambda = -2$

• Reasoning: The term $\frac{y-4}{0}$ in $L_1$'s equation implies that for any point on $L_1$, its $y$-coordinate must be $4$. Thus, for the point of intersection $B$, its $y$-coordinate must be $4$, which translates to $-2-3\lambda=4$, or $-6-3\lambda=0$. We can verify this $\lambda$ with the other terms: For $\lambda = -2$: $\frac{1+2(-2)}{1} = \frac{1-4}{1} = -3$ $\frac{3+6(-2)}{3} = \frac{3-12}{3} = \frac{-9}{3} = -3$ Since all parts are consistent, $\lambda = -2$ is the correct parameter value for the intersection point.

Step 4: Determine the Coordinates of Point $B$ Substitute the value of $\lambda = -2$ back into the parametric equations of $L_A$ (from Step 2) to find the coordinates of point $B$: $x_B = 7 + 2(-2) = 7 - 4 = 3$ $y_B = -2 - 3(-2) = -2 + 6 = 4$ $z_B = 11 + 6(-2) = 11 - 12 = -1$ So, the point of intersection $B$ is $(3,4,-1)$.

Step 5: Calculate the Distance $AB$ Finally, we calculate the distance between the given point $A(7,-2,11)$ and the intersection point $B(3,4,-1)$ using the 3D distance formula: $AB = \sqrt{(x_A-x_B)^2 + (y_A-y_B)^2 + (z_A-z_B)^2}$ $AB = \sqrt{(7-3)^2 + (-2-4)^2 + (11-(-1))^2}$ $AB = \sqrt{(4)^2 + (-6)^2 + (12)^2}$ $AB = \sqrt{16 + 36 + 144}$ $AB = \sqrt{196}$ $AB = 14$

Common Mistakes & Tips

• Misinterpretation of "along the line": Do not confuse this with finding the perpendicular distance from a point to a line. "Along the line" specifies the direction in which the distance is measured.
• Handling Zero Denominators in Line Equations: If a direction ratio is zero (e.g., $\frac{y-y_0}{0}$), it implies that the corresponding coordinate is constant for all points on the line (e.g., $y=y_0$). This forms a direct condition that helps in solving for the parameter.
• Careful with Arithmetic: Errors in squaring negative numbers or basic addition/subtraction can lead to incorrect final answers. Double-check calculations.

Summary

To find the distance of point $A$ from line $L_1$ along line $L_2$, we first constructed a new line $L_A$ passing through $A$ and parallel to $L_2$. We then found the point of intersection $B$ of this new line $L_A$ with the line $L_1$. The required distance is simply the Euclidean distance between points $A$ and $B$. By carefully using parametric forms of lines and handling the specific conditions arising from direction ratios, we determined the intersection point and subsequently the distance.

The final answer is $\boxed{\text{14}}$, which corresponds to option (D).