The point which lies on the plane passing through the points $A(\hat{i}-2\hat{j}-3\hat{k})$, $B(3\hat{i}-\hat{j}+4\hat{k})$, and $C(-3\hat{i}+2\hat{j}-5\hat{k})$ is:

Find the direction cosines of the unit vector perpendicular to the plane $\vec{r} \cdot (6 \hat{i} - 3 \hat{j} - 2 \hat{k}) + 1 = 0$.

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

If $(2, -1, 3)$ is the foot of the perpendicular drawn from the origin $(0, 0, 0)$ to a plane,then the equation of that plane is:

$A$ plane passing through the points $(0, -1, 0)$ and $(0, 0, 1)$ and making an angle $\frac{\pi}{4}$ with the plane $y - z + 5 = 0$ also passes through the point

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: