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Q.
Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x+y+z=3$. The feet of perpendiculars lie on the line
Three Dimensional Geometry
Solution:
Any point $B$ on line is $(2 \lambda-2,-\lambda-1,3 \lambda)$
Point $B$ lies on the plane for some $\lambda$.
$\Rightarrow (2 \lambda-2)+(-\lambda-1)+3 \lambda=3 $ or $ \lambda=3 / 2$
$\Rightarrow B \equiv(1,-5 / 2,9 / 2)$
The foot of the perpendicular from point $(-2,-1,0)$ on the plane is the point $A(0,1,2)$.
$\Rightarrow $ Direction ratio of $A B=\left(1, \frac{-7}{2}, \frac{5}{2}\right) \equiv(2,-7,5)$ Hence, feet of perpendicular lies on the line $\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5} .$