The straight line $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$ is

The vector equation of the line joining the points $i - 2j + k$ and $-2j + 3k$ is

If the line $\frac{2-x}{3}=\frac{3y-2}{4\lambda+1}=4-z$ makes a right angle with the line $\frac{x+3}{3\mu}=\frac{1-2y}{6}=\frac{5-z}{7}$,then $4\lambda+9\mu$ is equal to :

Let $P$ be the foot of the perpendicular from the point $Q(10,-3,-1)$ on the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}$. Then the area of the right-angled triangle $PQR$,where $R$ is the point $(3,-2,1)$,is

The distance of the line $3y - 2z - 1 = 0 = 3x - z + 4$ from the point $(2, -1, 6)$ is:

Show that the lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each other.