Let the acute angle bisector of the two planes $x-2y-2z+1=0$ and $2x-3y-6z+1=0$ be the plane $P$. Then which of the following points lies on $P$?

If $\alpha$ is the acute angle between the planes $P_1$ and $P_2$,where the combined equation of the planes $P_1$ and $P_2$ is $2x^2 - 6y^2 - 12z^2 + 18yz + 2zx + xy = 0$,then the value of $\cos \alpha$ is:

If $(2,3,-1)$ is the foot of the perpendicular from $(4,2,1)$ to a plane,then the equation of the plane is

The equation of a plane which passes through $(2, -3, 1)$ and is normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ is given by

The equation of the plane through the intersection of the planes $x + 2y + 3z - 4 = 0$ and $4x + 3y + 2z + 1 = 0$ and passing through the origin is:

Let $\pi_1$ be the plane passing through the point $2\hat{i}-\hat{j}+\hat{k}$ and perpendicular to the vector $a\hat{i}+2\hat{j}-3\hat{k}$,and $\pi_2$ be the plane passing through the point $\hat{i}+2\hat{j}-\hat{k}$ and perpendicular to the vector $\hat{i}-2\hat{j}+\hat{k}$. If $\theta$ is the angle between the planes $\pi_1$ and $\pi_2$ and $\cos \theta = -\sqrt{\frac{3}{7}}$,then the integral value of $a$ is: