The equation of 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:

$x+y+z=0$

Given point → P(0, 7, -7)

another line $\frac{x+1}{-3} = \frac{y-3}{2} = \frac{z+2}{1}$

so another point on place Q(-1, 3, -2)

vector on place $\vec{v_1} = -3\hat{i}+2\hat{j} + \hat{k}$

$\vec{PQ} = \vec{v_2} = -\hat{i}-4\hat{j} + 5\hat{k}$

so $\vec{n} = \vec{v_1} × \vec{v_2}$

$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 1 \\ -1 & -4 & 5\end{vmatrix}$

= $14\hat{i} + 14\hat{j} + 14\hat{k}$ is perpendicular to plane

So $\hat{i} + \hat{j} + \hat{k}$ is also perpendicular

⇒ $\vec{n} = \hat{i} + \hat{j} + \hat{k}$

so $(\vec{r} - \vec{a}) . \vec{n} = 0$ 

$\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$

$\vec{a} = 7\hat{j} - 7\hat{k}$

$\vec{r} . \vec{n} = \vec{a} . \vec{n}$

⇒  x + y + z = 0