To determine the equation of the plane \(P\), which is parallel to a given line and contains the line of intersection of two other planes, we will follow a series of logical steps.

Step 1: Determine the Normal of the Intersection Plane

The plane \(P\) contains the line of intersection of the given planes:

\(6x + 4y - 5z = 2\)

\(x - 2y + 3z = 0\)

The normals of these planes are \(\mathbf{n_1} = (6, 4, -5)\) and \(\mathbf{n_2} = (1, -2, 3)\).

Step 2: Find Direction Ratios of Intersection Line

The direction ratios of the line of intersection are given by the cross product \(\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2}\):

Calculating the determinant:

This gives the direction ratios as \((22, -23, -16)\).

Step 3: Plane Parallel to Given Line

Since plane \(P\) is parallel to a line with direction ratios \((1, 3, 2)\), the normal to the plane must be orthogonal to this line. Hence, we can set the normal vector of plane \(P\) as \(\mathbf{N} = a(1, 3, 2) + b\mathbf{d}\).

Step 4: Solve for Plane Equation

Since the plane passes through the line of intersection, substitute the general point on the line of intersection and ensure the plane equation holds.

We achieve this by setting:

Given options contain:

• \(2x+3y+2z-4 = 0\)
• \(2x-20y+29z-2 = 0\)
• \(2x-20y+29z+2 = 0\) (Correct Answer)
• \(x-3y+2z+5 = 0\)

On correctly solving for constants \(a\) and \(b\) such that the line lies on the plane, option \(2x-20y+29z+2 = 0\) satisfies all conditions.

Thus, the equation of the plane \(P\) is:

\(2x-20y+29z+2=0\).