The graph of the equation $y^2 + z^2 = 0$ in three-dimensional space is

Let a line $L$ be perpendicular to both the lines $L_1: \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}$ and $L_2: \frac{x-2}{1} = \frac{y-4}{4} = \frac{z-6}{7}$. If $\theta$ is the acute angle between the lines $L$ and $L_3: \frac{x-7}{2} = \frac{y-7}{1} = \frac{z}{2}$,then $\tan \theta$ is equal to:

The coordinates of the foot of the perpendicular drawn from the point $2 \hat{i} - \hat{j} + 5 \hat{k}$ to the line $\vec{r} = (11 \hat{i} - 2 \hat{j} - 8 \hat{k}) + \lambda(10 \hat{i} - 4 \hat{j} - 11 \hat{k})$ are

The lines $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}$ and $\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{2}$

The point of intersection of the lines represented by $r=(\hat{i}+2 \hat{j}-\hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$ and $r=(-\hat{i}-3 \hat{j}+7 \hat{k})+\mu(\hat{i}+2 \hat{j}-\hat{k})$ is

Let $M$ and $N$ be the feet of the perpendiculars drawn from the point $P(a, a, a)$ to the lines $L_1: x-y=0, z=1$ and $L_2: x+y=0, z=-1$ respectively. If $\angle MPN=90^{\circ}$,then $a^2=$