The reflection of the point $(-1, 3, 4)$ with respect to the plane $x - 2y = 0$ is .....

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

Let the line $\frac{x - 2}{3} = \frac{y - 1}{-5} = \frac{z + 2}{2}$ lie in the plane $x + 3y - \alpha z + \beta = 0$. Then $(\alpha, \beta)$ equals.

If the line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersects the plane $2x + 3y - z + 13 = 0$ at a point $P$ and the plane $3x + y + 4z = 16$ at a point $Q$,then $PQ$ is equal to

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:

The perpendicular distance from the origin to the plane containing the two lines $\frac{x+2}{3}=\frac{y-2}{5}=\frac{z+5}{7}$ and $\frac{x-1}{1}=\frac{y-4}{4}=\frac{z+4}{7}$ is: