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

The angle between the line of intersection of the two planes $r \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=5$ and $r \cdot(3 \hat{i}+3 \hat{j}-5 \hat{k})=3$,and the line $r=3 \hat{i}+2 \hat{j}+\hat{k}+t(5 \hat{i}+5 \hat{j}-7 \hat{k})$ is

The value of $\lambda$ for which the straight line $\frac{x-\lambda}{3}=\frac{y-1}{2+\lambda}=\frac{z-3}{-1}$ may lie on the plane $x-2y=0$ is

The equation of the plane passing through the origin and containing the line $\frac{x - 1}{5} = \frac{y - 2}{4} = \frac{z - 3}{5}$ is:

The point where the line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}$ meets the plane $2x+4y-z=1$ is:

$A$ plane $P$ contains the line of intersection of the planes $\vec{r} \cdot (\hat{i}+\hat{j}+\hat{k}) = 6$ and $\vec{r} \cdot (2\hat{i}+3\hat{j}+4\hat{k}) = -5$. If $P$ passes through the point $(0, 2, -2)$,then the square of the distance of the point $(12, 12, 18)$ from the plane $P$ is