The foot of the perpendicular from the point $A(1, 0, 3)$ to the line joining the points $B(4, 7, 1)$ and $C(3, 5, 3)$ 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

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

The equation of a line passing through the point $(2, 4, 6)$ and parallel to the line $3x + 4 = 4y - 1 = 1 - 4z$ is

The acute angle between the line joining the points $(2,1,-3)$ and $(-3,1,7)$ and a line parallel to $\frac{x-1}{3}=\frac{y}{4}=\frac{z+3}{5}$ is

If two lines $L_1$ and $L_2$ in space are defined by $L_1 = \{ x = \sqrt{\lambda} y + (\sqrt{\lambda} - 1), z = (\sqrt{\lambda} - 1)y + \sqrt{\lambda} \}$ and $L_2 = \{ x = \sqrt{\mu} y + (1 - \sqrt{\mu}), z = (1 - \sqrt{\mu})y + \sqrt{\mu} \}$,then $L_1$ is perpendicular to $L_2$ for all non-negative reals $\lambda$ and $\mu$ such that: