If the lines $\frac{x-1}{5}=\frac{y+1}{3}=\frac{3-z}{\lambda}$ and $\frac{x+1}{4}=\frac{1-3y}{15}=z+1$ are perpendicular to each other,then $\lambda=$

Statement $-1$: The point $A(1, 0, 7)$ is the mirror image of the point $B(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$.
Statement $-2$: The line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ bisects the line segment joining $A(1, 0, 7)$ and $B(1, 6, 3)$.

Two lines $L_1: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$ and $L_2: x=\alpha, \frac{y}{-1}=\frac{z}{2-\alpha}$ are coplanar. Then $\alpha$ can take value$(s)$.

The distance between the lines $\frac{x}{2}=\frac{y}{-1}=\frac{z}{2}$ and $\frac{x-1}{2}=\frac{y-1}{-1}=\frac{z-2}{2}$ is

If the coordinates of the points $A, B, C,$ and $D$ are $(1, 2, 3), (4, 5, 7), (-4, 3, -6),$ and $(2, 9, 2)$ respectively,then find the angle between the lines $AB$ and $CD$. (in $^\circ$)

Find the coordinates of the point where the line passing through the points $A(3, 4, 1)$ and $B(5, 1, 6)$ crosses the $XY$-plane.