Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plan $x+y+z=3$. The feet of perpendiculars lie on the line

$\text{ Let }\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}=k$

then $(2 k-2,-k-1,3 k)$ is a point on the line.

Let the foot of the perpendicular from this point on the

plane $x+y+z-3=0\text{ be }(\alpha ,\beta ,\gamma )$ then it is given by 

$\frac{\alpha -(2k-2)}{1}=\frac{\beta -(-k-1)}{1}=\frac{\gamma -3k}{1}$

$=-\frac{(2k-2)+(-k-1)+\left(3k\right)-3}{3}=\frac{6-4k}{3}$

Thus $\alpha =\frac{2k}{3},\beta =1-\frac{7k}{3},\gamma =2+\frac{5k}{3}$ then the line is given by

$\frac{x}{2/3}=\frac{y-1}{-7/3}=\frac{z-2}{5/3} \text{ ie. },\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$