The equation of the line passing through the point $(1, -3, 5)$ and making equal angles with the coordinate axes is:

The equation of the line passing through $(2, 3, 4)$ and parallel to the $Y$-axis is . . . . . . .

Let in a $\triangle ABC$,the length of the side $AC$ be $6$,the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle ABC$ is

For what value of $\lambda$ are the lines $\frac{x - 1}{1} = \frac{y - 2}{\lambda} = \frac{z + 1}{-1}$ and $\frac{x + 1}{-\lambda} = \frac{y + 1}{2} = \frac{z - 2}{1}$ perpendicular to each other?

If the shortest distance between the lines
$L_1: \overrightarrow{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, \lambda \in R$
$L_2: \overrightarrow{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, \mu \in R$
is $\frac{m}{\sqrt{n}}$,where $\operatorname{gcd}(m, n)=1$,then the value of $m+n$ equals.

If the mirror image of the point $P(3, 4, 9)$ in the line $\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$,then $14(\alpha+\beta+\gamma)$ is :