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 $^{\circ}$)

Let the plane $P$ pass through the intersection of the planes $2x + 3y - z = 2$ and $x + 2y + 3z = 6$,and be perpendicular to the plane $2x + y - z + 1 = 0$. If $d$ is the distance of $P$ from the point $(-7, 1, 1)$,then $d^2$ is equal to:

If the lines $L_1 : \frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $L_2 : \frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are perpendicular to each other, then the equation of a plane containing the line $L_1$ and parallel to the line $L_2$ for the value of $k$ that satisfies this condition is

If the distance between the plane $Ax - 2y + z = d$ and the plane containing the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}$ is $\sqrt{6}$,then $|d|$ is

Find the distance of the point $(1, -2, 3)$ from the plane $x - y + z = 5$ measured parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z - 1}{-6}$.

The point$(s)$ on the line $\vec r = \hat i + \hat j + \hat k + t(\hat i + 3\hat j - \hat k)$ at a distance of $3 \ units$ from the plane $\vec r \cdot (\hat i + 2\hat j + 2\hat k) + 2 = 0$ are