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

The given line is written in symmetric form: $$\frac{x+1}{-3} = \frac{y-3}{2} = \frac{z+2}{1} = \lambda$$.

From this form we read:
• One point on the line (take $$\lambda = 0$$) is $$A(-1,\,3,\,-2)$$.
• The direction ratios (d.r.’s) of the line are $$\langle -3,\,2,\,1\rangle$$.

The required plane must contain:
1. the entire line (hence the direction vector $$\langle -3,2,1\rangle$$ lies inside the plane), and
2. the external point $$B(0,\,7,\,-7)$$.

Let $$\vec{d} = \langle -3,2,1\rangle$$ be the direction vector of the line and $$\vec{AB}$$ be the vector joining the two known points in the plane.

Compute $$\vec{AB}:$$
$$\vec{AB} = \langle 0-(-1),\,7-3,\,-7-(-2)\rangle = \langle 1,4,-5\rangle.$$

A normal vector $$\vec{n}$$ to the plane is given by the cross-product of two non-parallel vectors lying in the plane: $$\vec{n} = \vec{d} \times \vec{AB}.$$

$$ \vec{n}= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ -3 & 2 & 1\\ 1 & 4 & -5 \end{vmatrix} = \mathbf{i}(2\cdot(-5)-1\cdot4) -\mathbf{j}((-3)(-5)-1\cdot1) +\mathbf{k}((-3)\cdot4-2\cdot1) = \langle -14,\,-14,\,-14\rangle. $$

Any non-zero scalar multiple serves as a normal, so we reduce to $$\vec{n} = \langle 1,1,1\rangle.$$

The plane through point $$A(-1,3,-2)$$ with normal $$\langle 1,1,1\rangle$$ is obtained from the point-normal form:
$$(x+1) + (y-3) + (z+2) = 0.$$

Simplifying, we get $$x + y + z = 0.$$

Verification: substitute the second given point $$B(0,7,-7)$$: $$0 + 7 + (-7) = 0$$, so the point lies on the plane. All conditions are satisfied.

Hence the required equation is
Option A which is: $$x + y + z = 0$$.