If the harmonic conjugate of $P(2, 3, 4)$ with respect to the line segment joining the points $A(3, -2, 2)$ and $B(6, -17, -4)$ is $Q(\alpha, \beta, \gamma)$,then $\alpha + \beta + \gamma =$

Let the values of $\lambda$ for which the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-\lambda}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\frac{1}{\sqrt{6}}$ be $\lambda_1$ and $\lambda_2$. Then the radius of the circle passing through the points $(0,0), (\lambda_1, \lambda_2)$ and $(\lambda_2, \lambda_1)$ is

$ABC$ is a triangle in a plane with vertices $A(2, 3, 5)$,$B(-1, 3, 2)$,and $C(\lambda, 5, \mu)$. If the median through $A$ is equally inclined to the coordinate axes,then the value of $\lambda + \mu$ is:

Find the shortest distance between the lines whose vector equations are $\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})$ and $\vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$.

Let $D$ be the foot of the perpendicular drawn from the point $A(2,0,3)$ to the line joining the points $B(0,4,1)$ and $C(-2,0,4)$. Then the ratio in which $D$ divides $BC$ is

The shortest distance between the lines $L_1: \bar{r} = \hat{i} + \hat{j} + \lambda(\hat{i} + \hat{j} - \hat{k})$ and $L_2: \bar{r} = \hat{j} + \hat{k} + \mu(\hat{j} + 2\hat{k} - \hat{i})$ is equal to: