Find the equation of the plane which is perpendicular to the plane `5x+3y+6z+8=0` adn which contains the line of intersection of the planes `x+2y+3z-4=0a n d2x+y-z+5=0.`

Explanation

To find the equation of the plane that is perpendicular to the plane \(5x + 3y + 6z + 8 = 0\) and contains the line of intersection of the planes \(x + 2y + 3z - 4 = 0\) and \(2x + y - z + 5 = 0\), we can follow these steps:

### Step 1: Identify the normal vector of the given plane
The normal vector of the plane \(5x + 3y + 6z + 8 = 0\) can be derived from the coefficients of \(x\), \(y\), and \(z\):
\[ \text{Normal vector} = (5, 3, 6) \]

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