The angle made by the vector $2\hat{i}-\hat{j}+\hat{k}$ with the plane represented by $\vec{r} \cdot(\hat{i}+\hat{j}+2\hat{k})=7$ is (in $^{\circ}$)

The centre of the circle given by $r \cdot (i + 2j + 2k) = 15$ and $|r - (j + 2k)| = 4$ is

The angle between the line $\frac{x - 1}{-2} = \frac{y - 2}{1} = \frac{z + 1}{2}$ and the plane $3x + 2y + 6z = 1$ is:

Let $PQR$ be a triangle with $R(-1, 4, 2)$. Suppose $M(2, 1, 2)$ is the midpoint of $PQ$. The distance of the centroid of $\triangle PQR$ from the point of intersection of the lines $\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1}$ and $\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1}$ is

The equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

Let the lines $l_1: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}$ and $l_2: 3x+2y+z-2=0=x-3y+2z-13$ be coplanar. If the point $P(a, b, c)$ on $l_1$ is nearest to the point $Q(-4, -3, 2)$,then $|a|+|b|+|c|$ is equal to