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(B) Let the point be $P = (0, 7, -7)$. The line is given by $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$.This line passes through point $A =

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$x + y + z = 0$

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(B) Let the point be $P = (0, 7, -7)$. The line is given by $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$.This line passes through point $A = (-1, 3, -2)$ and has a direction vector $\vec{b} = -3\hat{i} + 2\hat{j} + \hat{k}$.The vector $\vec{AP}$ connecting point $A$ to point $P$ is $\vec{AP} = (0 - (-1))\hat{i} + (7 - 3)\hat{j} + (-7 - (-2))\hat{k} = 1\hat{i} + 4\hat{j} - 5\hat{k}$.The normal vector $\vec{n}$ to the plane is the cross product of $\vec{AP}$ and $\vec{b}$:$\vec{n} = \vec{AP} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 4 & -5 \ -3 & 2 & 1 \end{vmatrix} = \hat{i}(4 - (-10)) - \hat{j}(1 - 15) + \hat{k}(2 - (-12)) = 14\hat{i} + 14\hat{j} + 14\hat{k}$.We can simplify the normal vector to $\vec{n}' = \hat{i} + \hat{j} + \hat{k}$.The equation of the plane is $(x - x_0)a + (y - y_0)b + (z - z_0)c = 0$,where $(x_0, y_0, z_0) = (0, 7, -7)$ and $(a, b, c) = (1, 1, 1)$.$1(x - 0) + 1(y - 7) + 1(z + 7) = 0 \implies x + y + z = 0$.

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

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