If the points $A(-1, 3, 2)$,$B(-4, 2, -2)$,and $C(5, 5, \lambda)$ are collinear,then $\lambda = $

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

Let $A(2,3,5), B(-1,3,2), C(\lambda, 5, \mu)$ be the vertices of $\triangle ABC$. If the median through the vertex $A$ is equally inclined to the coordinate axes,then

The length of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$ is (in $\text{ units}$)

If the distances of the point $P(1, 2, a)$ from the line $L: \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ along the lines $L_{1}: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$ and $L_{2}: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$ are equal, then $a+b+c$ is equal to

The shortest distance between the skew lines $\frac{x-2}{1}=\frac{y-3}{-2}=\frac{z+5}{1}$ and $\frac{x-1}{-1}=\frac{y+2}{3}=\frac{z-4}{2}$ is