The equation of the planes passing through the line of intersection of the planes $3x - y - 4z = 0$ and $x + 3y + 6 = 0$ whose distance from the origin is $1$,are

The line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta=0$,then the value of $\alpha^2+\alpha\beta+\beta^2$ is

Let $P$ be the plane containing the straight line $\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$ and perpendicular to the plane containing the straight lines $\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$ and $\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$. If $d$ is the distance of $P$ from the point $(2,-5,11)$,then $d^{2}$ is equal to.

The equation of the plane $\pi$ through the line of intersection of the planes $\pi_1 \equiv x+3y-6=0$ and $\pi_2 \equiv 3x-y+4z=0$ is $\pi_1+\lambda \pi_2=0$. If the plane $\pi$ is at unit distance from the origin,then an equation of the plane $\pi$ is

What is the distance of the point $(2, 3, 4)$ from the plane $3x - 6y + 2z + 11 = 0$?

The angle $\theta$ between the line $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})$ and the plane $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 8$ is