The equation of the plane passing through the point $(1,1,1)$ and perpendicular to the planes $2x-y-2z=5$ and $3x-6y+2z=7$ is

Let a plane $P$ contain the points $\hat{i}, \hat{j}$ and $\hat{i}+\hat{j}+\hat{k}$. Let $L$ be the line through the point $A(3, 0, -5)$ and parallel to the vector $\hat{i}-\hat{j}+\hat{k}$. The equation of the normal to the plane $P$ passing through point $A$ is:

$A$ tetrahedron has vertices $P(1,2,1), Q(2,1,3), R(-1,1,2)$ and $O(0,0,0)$. Then the angle between the faces $OPQ$ and $PQR$ is

If the length of the perpendicular drawn from the origin to a plane is $7$ units and its direction ratios are $-3, 2, 6$,then the equation of the plane is:

The equation of the plane passing through $(1,0,0)$ and $(0,1,0)$ and making an angle of $45^{\circ}$ with the plane $x+y-3=0$ 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: