Perpendiculars are drawn from the 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 the perpendiculars lie on the line

$\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$

The coordinates of any point on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ are given by  $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}= r$.

So, the coordinates of any point on the given line are $P(2r - 2, -r -1, 3r)$.

Let M $(u, v, w)$ be the feet of perpendicular drawn from P on the plane x + y + z = 3. Then, the direction ratios of PM are proportional to $2r -2 - u, -r -1 - v, 3r - w.$ But, PM is normal to the plane the direction ratios of whose normal are proportional to 1, 1, 1.

$∴ \frac{2r-2-u}{1}=\frac{-r-1-v}{1}=\frac{3r-w}{1}= λ (say)$

$⇒ u = 2r - 2- λ, v = -r - 1 - λ, w= 3r - λ $

Since $M(u, v, w)$ lies on the plane $ x + y + z = 3.$ Therefore,

$u + v + w = 3 $

$⇒ 2r - 2 - λ - r - 1 - λ + 3r - λ = 3 $

$⇒ 4r - 3 - 3λ =3 ⇒λ= \frac{4r-6}{3}$

$∴ u = 2r - 2 - \frac{4r-6}{3}, v = -r -1 -\frac{4r-6}{3}, w = 3r -\frac{4r-6}{3}$

$⇒ u = \frac{2r}{3}, v = \frac{-7r+3}{3}, w = \frac{5r+6}{3}$

$⇒ \frac{u}{2/3}=\frac{v-1}{-7/3}=\frac{w-2}{5/3}= r$

Hence, $(u, v, w)$ lies on the line

$\frac{x}{2/3}= \frac{y-1}{-7/3}=\frac{w-2}{5/3} \,$ or  $\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$