The equation of a plane containing the line of intersection of the planes $2x - y - 4 = 0$ and $y + 2z - 4 = 0$ and passing through the point $(1, 1, 0)$ is

The $XY$-plane divides the line segment joining the points $A(2, 3, -5)$ and $B(-1, -2, -3)$ in the ratio:

Let the line $\ell: x = \frac{1-y}{-2} = \frac{z-3}{\lambda}, \lambda \in R$ meet the plane $P: x + 2y + 3z = 4$ at the point $(\alpha, \beta, \gamma)$. If the angle between the line $\ell$ and the plane $P$ is $\cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$,then $\alpha + 2\beta + 6\gamma$ is equal to

$A$ plane $\pi_1$ contains the vectors $\bar{i}+\bar{j}$ and $\bar{i}+2\bar{j}$. Another plane $\pi_2$ contains the vectors $2\bar{i}-\bar{j}$ and $3\bar{i}+2\bar{k}$. $\bar{a}$ is a vector parallel to the line of intersection of $\pi_1$ and $\pi_2$. If the angle $\theta$ between $\bar{a}$ and $\bar{i}-2\bar{j}+2\bar{k}$ is acute, then $\theta=$

Equation of the plane which passes through the point of intersection of lines $\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ and has the largest distance from the origin is

The distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is