Key Concepts and Formulas

• Equation of a Line in 3D (Parametric Form): A line passing through a point $(x_0, y_0, z_0)$ and having a direction vector $\vec{d} = (a, b, c)$ can be represented as: $x = x_0 + a\lambda$ $y = y_0 + b\lambda$ $z = z_0 + c\lambda$ where $\lambda$ is a scalar parameter. The symmetric form is $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$.
• Intersection of Two Lines: To find the intersection point of two lines, express points on each line using different parametric variables (e.g., $\lambda$ and $\mu$). Equate the corresponding $x, y, z$ coordinates to form a system of linear equations. Solving this system for the parameters will give the coordinates of the intersection point.
• Distance between Two Points in 3D: The distance $D$ between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is given by the distance formula: $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

Step-by-Step Solution

Step 1: Formulate the Equation of the First Line (L1)

• What we are doing: We need to find the parametric equation of the line L1.
• Why we are doing it: This form allows us to represent any point on L1 in terms of a single parameter, which is essential for finding the intersection point later.
• Working:
  1. Identify the given point: L1 passes through $(x_0, y_0, z_0) = (-1, 2, 1)$.
  2. Identify the direction vector: L1 is parallel to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$. Since parallel lines share the same direction vector (or proportional ones), we can directly take the direction ratios from the given line. The direction vector is $\vec{d}_1 = (2, 3, 4)$.
  3. Write the parametric equations for L1: Using the point $(-1, 2, 1)$ and direction vector $(2, 3, 4)$ with parameter $\lambda$: $x = -1 + 2\lambda \quad (L1_x)$ $y = 2 + 3\lambda \quad (L1_y)$ $z = 1 + 4\lambda \quad (L1_z)$ Any point on L1 can be represented as $P_1(-1+2\lambda, 2+3\lambda, 1+4\lambda)$.

Step 2: Find the Intersection Point P of L1 and L2

• What we are doing: We will find the coordinates of the point P where line L1 intersects the second line L2.
• Why we are doing it: The problem asks for the distance from this intersection point P to another point Q.
• Working:
  1. Write the parametric equations for L2: The second line (L2) is given by $\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}$.
     • A point on L2 is $(-2, 3, 4)$.
     • The direction vector for L2 is $\vec{d}_2 = (3, 2, 1)$.
     • Parametric equations for any point on L2 (using a different parameter, $\mu$): $x = -2 + 3\mu \quad (L2_x)$ $y = 3 + 2\mu \quad (L2_y)$ $z = 4 + \mu \quad (L2_z)$ Any point on L2 can be represented as $P_2(-2+3\mu, 3+2\mu, 4+\mu)$.
  2. Equate the coordinates of L1 and L2 to find P: If the lines intersect at point P, their coordinates must be equal for specific values of $\lambda$ and $\mu$. $-1 + 2\lambda = -2 + 3\mu \quad \Rightarrow \quad 2\lambda - 3\mu = -1 \quad (1)$ $2 + 3\lambda = 3 + 2\mu \quad \Rightarrow \quad 3\lambda - 2\mu = 1 \quad (2)$ $1 + 4\lambda = 4 + \mu \quad \Rightarrow \quad 4\lambda - \mu = 3 \quad (3)$
  3. Solve the system of linear equations: We can solve this system for $\lambda$ and $\mu$. From equation (3), express $\mu$ in terms of $\lambda$: $\mu = 4\lambda - 3$ Substitute this expression for $\mu$ into equation (1): $2\lambda - 3(4\lambda - 3) = -1$ $2\lambda - 12\lambda + 9 = -1$ $-10\lambda = -10$ $\lambda = 1$ Now, substitute $\lambda = 1$ back into the expression for $\mu$: $\mu = 4(1) - 3 = 1$
  4. Verify the parameters: It's good practice to check if these values of $\lambda$ and $\mu$ satisfy the remaining equation (2): $3\lambda - 2\mu = 3(1) - 2(1) = 3 - 2 = 1$ Since $1=1$, the values $\lambda=1$ and $\mu=1$ are consistent, confirming that the lines intersect.
  5. Find the coordinates of P: Substitute $\lambda=1$ into the parametric equations for L1 (or $\mu=1$ into L2's equations; both will yield the same point): $x_P = -1 + 2(1) = 1$ $y_P = 2 + 3(1) = 5$ $z_P = 1 + 4(1) = 5$ Thus, the intersection point is $P(1, 5, 5)$.

Step 3: Calculate the Distance between Point P and Point Q

• What we are doing: We will use the distance formula to find the distance between the intersection point P and the given point Q.
• Why we are doing it: This is the final requirement of the problem statement.
• Working:
  1. Identify the coordinates:
     • $P(x_1, y_1, z_1) = (1, 5, 5)$
     • $Q(x_2, y_2, z_2) = (4, -5, 1)$
  2. Apply the distance formula: $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$ $D = \sqrt{(4-1)^2 + (-5-5)^2 + (1-5)^2}$ $D = \sqrt{(3)^2 + (-10)^2 + (-4)^2}$ $D = \sqrt{9 + 100 + 16}$ $D = \sqrt{125}$
  3. Simplify the radical: $D = \sqrt{25 \times 5}$ $D = 5\sqrt{5}$

Common Mistakes & Tips

• Sign Errors: Be very careful with signs when setting up line equations (e.g., $x+2$ implies $x_0=-2$) and when applying the distance formula.
• Algebraic Accuracy: Solving the system of three linear equations for $\lambda$ and $\mu$ is a common source of errors. Double-check your calculations, especially during substitution.
• Parameter Consistency: Always verify the calculated parameter values ($\lambda$ and $\mu$) in all three equations to ensure consistency. If they don't satisfy all three, the lines either don't intersect or there's an error in your calculation.

Summary

This problem involved a systematic application of 3D geometry principles. First, we established the parametric equation for the first line using its given point and direction vector. Next, we found the intersection point P by setting the parametric coordinates of the two lines equal and solving the resulting system of linear equations. Finally, we calculated the distance between this intersection point P and the given point Q using the standard 3D distance formula. The intersection point was found to be $P(1,5,5)$, and the distance to $Q(4,-5,1)$ was $5\sqrt{5}$.

The final answer is $\boxed{5\sqrt{5}}$, which corresponds to option (C).