The square of the distance of the point of intersection of the line x-12=y-23=z+16 and the…

Question

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

Answer: 61

Accepted Answer

Points on the line: $(2\lambda+1,3\lambda+2,6\lambda-1)$. Substituting into the plane:
$$2(2\lambda+1)-(3\lambda+2)+(6\lambda-1)=6\implies 7\lambda-1=6\implies\lambda=1.$$
Point of intersection: $P(3,5,5)$.
$PQ^2=(3+1)^2+(5+1)^2+(5-2)^2=16+36+9=61.$

Solutions · 1