The shortest distance between the lines $\frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ is equal to -

Show that the line through the points $(1, -1, 2)$ and $(3, 4, -2)$ is perpendicular to the line through the points $(0, 3, 2)$ and $(3, 5, 6)$.

If the image of the point $(4,4,3)$ in the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to

If the lines $\frac{2x-4}{\lambda} = \frac{y-1}{2} = \frac{z-3}{1}$ and $\frac{x-1}{1} = \frac{3y-1}{\lambda} = \frac{z-2}{1}$ are perpendicular to each other,then $\lambda=$

Let a line passing through the point $P(4,1,0)$ intersect the line $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $L_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then $\left|\begin{array}{lll}1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{array}\right|$ is equal to

The perpendicular distance of the line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}$ from the point $P(2,-10,1)$ is: