Step 1: Gather data from the line.
The line $\dfrac{x+1}{-3} = \dfrac{y-3}{2} = \dfrac{z+2}{1}$ passes through $A(-1, 3, -2)$ with direction $\bar{u} = (-3, 2, 1)$.
Step 2: Use the extra point.
The plane also passes through $B(0, 7, -7)$. The vector along the plane from $A$ to $B$ is
\[ \overline{AB} = (0+1,\ 7-3,\ -7+2) = (1, 4, -5) \]
Step 3: Get the normal.
The normal $\bar{n}$ is perpendicular to both $\bar{u}$ and $\overline{AB}$, so it is their cross product.
Step 4: Compute the cross product.
\[ \bar{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 1 \\ 1 & 4 & -5 \end{vmatrix} = \hat{i}(-10 - 4) - \hat{j}(15 - 1) + \hat{k}(-12 - 2) \]
\[ = -14\hat{i} - 14\hat{j} - 14\hat{k} \]
Step 5: Simplify the normal.
Divide by $-14$: $\bar{n} = (1, 1, 1)$.
Step 6: Write the plane.
Using point $B(0,7,-7)$:
\[ 1(x-0) + 1(y-7) + 1(z+7) = 0 \;\Rightarrow\; x + y + z = 0 \]
\[ \boxed{x + y + z = 0} \]