The equation of the plane passing through the point (0, 7, -7) and containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$, is

$x+ y + z= 0 $

Let the equation of a plane passing through (0,7,-7) be

$a(x-0) + b(y-7) + c(z+7) = 0 $ ........(i)

The line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$passes through the point (-1, 3, -2) and its direction ratios are proportional to -3, 2, 1.

If (i) contains this line, it must pass through (-1, 3, -2) and must be parallel to the line. Therefore,

$a(-1) + b(-4) + c(5) = 0 $ .......(ii)

and, $-3a +  2b + 1c = 0 $ ..........(iii)

On solving (ii) and (iii) by cross-multiplication, we get

$\frac{a}{-14}=\frac{b}{-14}=\frac{c}{-14}$

$⇒ \frac{a}{1}=\frac{b}{1}=\frac{c}{1}=λ$ (say) ⇒ $a = λ, b = λ, c = λ.$

Putting the values of a, b, c in (i), we obtain

$λ(x-0) + λ (y-7) + λ (z + 7) = 0 ⇒ x + y + z = 0.$