The Cartesian equation of the line passing through the point $(-1, 3, -2)$ and perpendicular to the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5}$ is

Points $P$ and $Q$ are given by $\vec{OP} = \hat{i} - \hat{j} - \hat{k}$ and $\vec{OQ} = -\hat{i} + \hat{j} + \hat{k}$. $A$ line along the vector $\vec{a} = \hat{i} + \hat{j}$ passes through the point $P$ and another line along the vector $\vec{b} = \hat{j} - \hat{k}$ passes through the point $Q$. If a line along the vector $\vec{c} = \hat{i} - \hat{j} + \hat{k}$ intersects both the lines along the vectors $\vec{a}$ and $\vec{b}$ at $L$ and $M$ respectively,then $\vec{PM} =$

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

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

$A$ line with direction cosines proportional to $2, 1, 2$ meets the line $L_1$ passing through $(0, -1, 0)$ with direction ratios $1, 1, 1$ at $A(x, y, z)$ and another line $L_2$ at $B(1, 1, 1)$. Then $x+y+z=$

The point of intersection of the lines $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ and $\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}$ is