If the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane $\ell x+m y-z=9$,then $\ell^2+m^2$ is

The direction cosines of the line formed by the intersection of the planes $x - y + 2z = 5$ and $3x + y + z = 6$ are:

If the points $(1, 1, \lambda)$ and $(-3, 0, 1)$ are equidistant from the plane $3x + 4y - 12z + 13 = 0$,then the values of $\lambda$ are

Find the equation of the plane through the intersection of the planes $\vec{r} \cdot(\hat{i}+3 \hat{j})-6=0$ and $\vec{r} \cdot(3 \hat{i}-\hat{j}-4 \hat{k})=0,$ whose perpendicular distance from the origin is unity.

The distance of the point of intersection of the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ and the plane $x + y + z = 17$ from the point $(3, 4, 5)$ is given by

Find the equation of the plane containing the lines $\vec{r} = (\hat{i} + \hat{j}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})$ and $\vec{r} = (\hat{i} + \hat{j}) + \mu(-\hat{i} + \hat{j} - 2\hat{k})$.