The angle $\theta$ between the line $x - 1 = \frac{y - 2}{-1} = \frac{z - 3}{2}$ and the plane $\vec{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 10$ is

The direction ratios of the line of intersection of the planes $x-y+z-5=0$ and $x-3y-6=0$ are:

The symmetric equation of the line formed by the intersection of the planes $3x + 2y + z - 5 = 0$ and $x + y - 2z - 3 = 0$ is:

Let $\pi_1$ be a plane passing through the point $\hat{i}+\hat{j}+\hat{k}$ and perpendicular to the vector $-\hat{j}+2\hat{k}$. Let the line $L$ passing through the points $3\hat{i}-2\hat{j}+\hat{k}$ and $-\hat{i}+3\hat{j}+\hat{k}$ be a normal to the plane $\pi_2$. If the angle between the planes $\pi_1$ and $\pi_2$ is $\theta$, then $\cos \theta =$

Let $Q$ be the foot of the perpendicular drawn from the point $P(1, 2, 3)$ to the plane $x + 2y + z = 14$. If $R$ is a point on the plane such that $\angle PRQ = 60^{\circ}$,then the area of $\triangle PQR$ is equal to:

Find the distance of the point $2\hat{i} + \hat{j} - \hat{k}$ from the plane $\vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) = 9$.