Find the vector equation of tha plane which passes through the point (3, 2, -3) and perpendicular
to a line with direction ratios (1, 2, -1).

Explanation

To find the vector equation of the plane that passes through the point (3, 2, -3) and is perpendicular to a line with direction ratios (1, 2, -1), we can follow these steps:

### Step 1: Identify the Point and Normal Vector
The point through which the plane passes is given as \( P(3, 2, -3) \). The direction ratios of the line, which also serve as the normal vector \( \mathbf{n} \) to the plane, are given as \( (1, 2, -1) \). Therefore, we can express the normal vector as:
\[ \mathbf{n} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \]

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