Find the feet of perpendiculars on a plane

Question

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
$(A)\quad\cfrac{x}{5}=\cfrac{y-1}{8}=\cfrac{z-2}{-13}\hspace{1cm}$

$(B)\quad\cfrac{x}{2}=\cfrac{y-1}{3}=\cfrac{z-2}{-5}\hspace{1cm}$

$(C)\quad\cfrac{x}{4}=\cfrac{y-1}{3}=\cfrac{z-2}{-7}\hspace{1cm}$

$(D)\quad\cfrac{x}{2}=\cfrac{y-1}{-7}=\cfrac{z-2}{5}$

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I have no idea how to solve this question. Please help me. Thanks.

Answer 1

HINT: Firstly, you find the plane which contains the line and orthogonal the plane.

After that, you can compute the intersection between two planes.