The square of the distance of the point of intersection

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Question:

The square of the distance of the point of intersection of the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}$ and the plane $2 \mathrm{x}-\mathrm{y}+\mathrm{z}=6$ from the point $(-1,-1,2)$ is______.

Solution:

$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}=\lambda$

$x=2 \lambda+1, y=3 \lambda+2, z=6 \lambda-1$

for point of intersection of line \& plane

$2(2 \lambda+1)-(3 \lambda+2)+(6 \lambda-1)=6$

$7 \lambda=7 \Rightarrow \lambda=1$

point: $(3,5,5)$

$(\text { distance })^{2}=(3+1)^{2}+(5+1)^{2}+(5-2)^{2}$

$=16+36+9=61$

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