The distance of the point $O(\vec{0})$ from the plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=5$ measured parallel to the vector $2 \hat{i}+3 \hat{j}-6 \hat{k}$ is:

Let $P(x_1, y_1, z_1)$ be the foot of the perpendicular drawn from the point $Q(2, -2, 1)$ to the plane $x - 2y + z = 1$. If $d$ is the perpendicular distance from the point $Q$ to the plane and $l = x_1 + y_1 + z_1$,then the value of $l + 3d^2$ is:

Show that the lines $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$ and $\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$ are coplanar.

Find the equation of the plane passing through the intersection of the planes $x + y + z = 6$ and $2x + 3y + 4z + 5 = 0$ and the point $(1, 1, 1)$.

If the angle between the line $x = \frac{y - 1}{2} = \frac{z - 3}{\lambda}$ and the plane $x + 2y + 3z = 4$ is $\cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$,then $\lambda = \dots$

Let the foot of the perpendicular from the point $A(4, 3, 1)$ on the plane $P: x - y + 2z + 3 = 0$ be $N$. If $B(5, \alpha, \beta)$,where $\alpha, \beta \in \mathbb{Z}$,is a point on the plane $P$ such that the area of the triangle $ABN$ is $3\sqrt{2}$,then $\alpha^2 + \beta^2 + \alpha\beta$ is equal to $...........$.