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

If the straight lines $x = 1 + s, y = -3 - \lambda s, z = 1 + \lambda s$ and $x = t/2, y = 1 + t, z = 2 - t$,with parameters $s$ and $t$ respectively,are coplanar,then $\lambda$ equals

The equation of the plane passing through the intersection of the planes $2x - 5y + z = 3$ and $x + y + 4z = 5$ and parallel to the plane $x + 3y + 6z = 1$ is $x + 3y + 6z = k$,where $k$ is:

If the plane $4x + 4y - kz = 0$ is the equation of the plane passing through the origin and containing the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$,find the value of $k$.

Consider the line $L$ given by the equation $\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1}$. Let $Q$ be the mirror image of the point $P_0(2,3,-1)$ with respect to $L$. Let a plane $P$ be such that it passes through $Q$,and the line $L$ is perpendicular to $P$. Then which of the following points is on the plane $P$?

For real numbers $\alpha$ and $\beta \neq 0$,if the point of intersection of the straight lines $\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3}$ and $\frac{x-4}{\beta}=\frac{y-6}{3}=\frac{z-7}{3}$ lies on the plane $x+2y-z=8$,then $\alpha-\beta$ is equal to: