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

Explanation

To find the vector equation of the line passing through the point \( (1, -1, 2) \) and perpendicular to the plane given by the equation \( 2x - y + 3z - 5 = 0 \), we will follow these steps:

### Step 1: Identify the normal vector of the plane
The equation of the plane is given in the form \( Ax + By + Cz + D = 0 \). From the equation \( 2x - y + 3z - 5 = 0 \), we can identify the coefficients:
- \( A = 2 \)
- \( B = -1 \)
- \( C = 3 \)

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