Find the equation of the plane passing through the point `( 1, 3, 2)`and perpendicular to each of the planes `x + 2y + 3z = 5`and `3x + 3y + z = 0`.

Explanation

To find the equation of the plane that passes through the point \( (1, 3, 2) \) and is perpendicular to the planes given by the equations \( x + 2y + 3z = 5 \) and \( 3x + 3y + z = 0 \), we can follow these steps:

### Step 1: Identify the normals of the given planes
The normal vector of a plane can be derived from the coefficients of \( x \), \( y \), and \( z \) in its equation.

For the first plane \( x + 2y + 3z = 5 \):
- The normal vector \( \mathbf{n_1} = (1, 2, 3) \).

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