We need to find the equation of a plane that passes through the point $(-1, -2, 3)$ and is perpendicular to the planes $x - 3y + 2z = 7$ and $2x - 2y - z = -3$.

1. Problem Description

We need to find the equation of a plane that passes through the point $(-1, -2, 3)$ and is perpendicular to the planes $x - 3y + 2z = 7$ and $2x - 2y - z = -3$.

2. Solution Steps

Since the plane we are looking for is perpendicular to both given planes, its normal vector is parallel to the cross product of the normal vectors of the given planes.

First, we find the normal vectors of the given planes:

The normal vector of the plane $x - 3y + 2z = 7$ is $\vec{n_1} = <1, -3, 2>$.

The normal vector of the plane $2x - 2y - z = -3$ is $\vec{n_2} = <2, -2, -1>$.

Next, we compute the cross product of these two normal vectors:

$\vec{n} = \vec{n_1} \times \vec{n_2} = <1, -3, 2> \times <2, -2, -1>$

$= <(-3)(-1) - (2)(-2), (2)(2) - (1)(-1), (1)(-2) - (-3)(2)>$

$= <3 + 4, 4 + 1, -2 + 6>$

$= <7, 5, 4>$

This vector $\vec{n} = <7, 5, 4>$ is the normal vector to the plane we are trying to find.

The general equation of a plane with normal vector $<a, b, c>$ is given by:

$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$

where $(x_0, y_0, z_0)$ is a point on the plane.

We are given the point $(-1, -2, 3)$, so $x_0 = -1$, $y_0 = -2$, and $z_0 = 3$.

Using the normal vector $<7, 5, 4>$ and the point $(-1, -2, 3)$, we get the equation:

$7(x - (-1)) + 5(y - (-2)) + 4(z - 3) = 0$

$7(x + 1) + 5(y + 2) + 4(z - 3) = 0$

$7x + 7 + 5y + 10 + 4z - 12 = 0$

$7x + 5y + 4z + 5 = 0$

$7x + 5y + 4z = -5$

3. Final Answer

The equation of the plane is $7x + 5y + 4z = -5$.