Find the vector and cartesian equation of plane which passes through the point `(1,3,-2)` and normal to the vector `(2hati+hatj-2hatk)` .

Explanation

To find the vector and Cartesian equation of the plane that passes through the point \( (1, 3, -2) \) and is normal to the vector \( (2\hat{i} + \hat{j} - 2\hat{k}) \), we can follow these steps:

### Step 1: Identify the Point and Normal Vector
The point through which the plane passes is given as \( (1, 3, -2) \). We can denote this point as \( A(1, 3, -2) \). The normal vector to the plane is given as \( \vec{n} = 2\hat{i} + \hat{j} - 2\hat{k} \).

### Step 2: Write the Position Vector of the Point
The position vector \( \vec{A} \) of the point \( A \) can be expressed as:
\[
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