If the distance of point of intersection of lines $\frac{x-4}{1} = \frac{y+3}{-4} = \frac{z+1}{7}$ and $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z+10}{8}$ from $(1, -4, 7)$ is $a$, then $\frac{a^2}{13}$ is equal to __________.
Step-by-Step Solution
Key Concept: Find the intersection of two lines by solving parametric equations simultaneously, then calculate the distance to a given point.
From $4 + \lambda_1 = 1 + 2\lambda_2$, we get $\lambda_1 - 2\lambda_2 = -3$. From $-3 - 4\lambda_1 = -1 - 3\lambda_2$, we get $4\lambda_1 - 3\lambda_2 = -2$. Solving these two equations: $-5 \times 2 = -10$ yields $\lambda_2 = 2$ and $\lambda_1 = 1$. The point of intersection is $(4 + 1, -3 - 4, -1 + 7) = (5, -7, 6)$. The distance from this point to $(1, -4, 7)$ is $\sqrt{16 + 9 + 1} = \sqrt{26}$.
Correct Answer: Let me verify the solution step by step.
**Finding the point of intersection:**
Line 1: $\frac{x-4}{1} = \frac{y+3}{-4} = \frac{z+1}{7} = \lambda_1$
Parametric form: $(x,y,z) = (4+\