For the following two (02) items: A plane P is parallel to the line...

The equation of the plane passing through the line of intersection of the given planes is

$(6x + 4y - 5z - 2) + \lambda(x - 2y + 3z) = 0$

$(6 + \lambda)x + (4 - 2\lambda)y + (-5 + 3\lambda)z - 2 = 0$

Since this plane is parallel to the line with direction ratios $\langle 1, 3, 2 \rangle$, the normal to the plane is perpendicular to the line. Thus,

$1(6 + \lambda) + 3(4 - 2\lambda) + 2(-5 + 3\lambda) = 0$

$6 + \lambda + 12 - 6\lambda - 10 + 6\lambda = 0$

$\lambda + 8 = 0 \Rightarrow \lambda = -8$

Substituting $\lambda = -8$ in the equation of the plane, we get

$(6 - 8)x + (4 - 2(-8))y + (-5 + 3(-8))z - 2 = 0$

$-2x + 20y - 29z - 2 = 0$

$2x - 20y + 29z + 2 = 0$

Answer: $2x - 20y + 29z + 2 = 0$