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

The ratio in which the line segment joining the points $(2, -4, 3)$ and $(-4, 5, -6)$ is divided by the plane $3x + 2y + z - 4 = 0$ is:

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

If the line $\frac{x+1}{1}=\frac{y-k}{11}=\frac{z-4}{-5}$ lies in the plane $2x+py+7z-41=0$ which is perpendicular to the plane $x+4y-2z+13=0$,then $k=$

Let $P$ be the plane,which contains the line of intersection of the planes $x + y + z - 6 = 0$ and $2x + 3y + z + 5 = 0$ and it is perpendicular to the $xy$-plane. Then the distance of the point $(0, 0, 256)$ from $P$ is equal to

The vector equation of a plane passing through the line of intersection of the planes $\overline{r} \cdot(\overline{i}-2 \overline{k})=3$ and $\overline{r} \cdot(2 \overline{j}+\overline{k})=5$,and passing through the point $\overline{i}+2 \overline{j}+3 \overline{k}$,is: