The equation of the plane containing the straight line $\frac{x}{3}=\frac{y}{2}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{4}=\frac{y}{3}=\frac{z}{2}$ and $\frac{x}{2}=\frac{y}{-4}=\frac{z}{3}$ is:

Let $A(2,5,7)$ be the image of the point $B(1,-2,3)$ with respect to a plane $\pi$. Let $C$ be the point where $AB$ meets the plane $\pi$. Let $D=(2,1,6)$. Then the direction cosines of $CD$ are

The plane $2x - y + z = 4$ intersects the line segment joining the points $A(a, -2, 4)$ and $B(2, b, -3)$ at the point $C$ in the ratio $2:1$. The distance of the point $C$ from the origin is $\sqrt{5}$. If $ab < 0$ and $P$ is the point $(a - b, b, 2b - a)$,then $CP^2$ is equal to:

The equation of the plane passing through the point $(1,1,1)$ and through the line of intersection of $x+2y-z+1=0$ and $3x-y-4z+3=0$ is

If the line joining the points $\overline{i} + 2\overline{j}$ and $\overline{j} - 2\overline{k}$ intersects the plane passing through the points $2\overline{i} - \overline{j}$,$2\overline{j} + 3\overline{k}$,and $\overline{k} - 2\overline{i}$ at $\overline{r}$,then $\overline{r} \cdot (\overline{i} + \overline{j} + \overline{k}) = $

The line $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{0}$ is parallel to