If the lines $\frac{3-x}{2}=\frac{5y-2}{3\lambda+1}=5-z$ and $\frac{x+2}{-1}=\frac{1-3y}{7}=\frac{4-z}{2\mu}$ are at right angles,then $7\lambda-10\mu=$

The Cartesian equation of a line is $2x - 3 = 3y + 1 = 5 - 6z$. The vector equation of the line passing through the point $(7, -5, 0)$ and parallel to the given line is

$\triangle ABC$ is formed by $A(1, 8, 4)$,$B(0, -11, 4)$,and $C(2, -3, 1)$. If $D$ is the foot of the perpendicular drawn from $A$ to $BC$,then the coordinates of $D$ are

Find the acute angle between the line joining the points $(2, 1, -3)$ and $(-3, 1, 7)$ and the line parallel to $\frac{x - 1}{3} = \frac{y}{4} = \frac{z + 3}{5}$ passing through the point $(-1, 0, 4)$.

If the image of the point $P(a, 2, a)$ in the line $\frac{x}{2} = \frac{y+a}{1} = \frac{z}{1}$ is $Q$ and the image of $Q$ in the line $\frac{x-2b}{2} = \frac{y-a}{1} = \frac{z+2b}{-5}$ is $P$, then $a+b$ is equal to:

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is