Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=z$ at a distance $4 \sqrt{14}$ from the point $(1,-1,0)$ and nearer to the origin. Then the shortest distance, between the lines $\frac{x-\alpha}{1}=\frac{y-\beta}{2}=\frac{z-\gamma}{3}$ and $\frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1}$, is equal to

If the image of the point $\mathrm{P}(1,2, a)$ in the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{7-\mathrm{z}}{2}$ is $\mathrm{Q}(5, b, \mathrm{c})$, then $a^2+b^2+c^2$ is equal to

Let the line L pass through the point $(-3, 5, 2)$ and make equal angles with the positive coordinate axes. If the distance of L from the point $(-2, r, 1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of $r$ is :

Let the line $L_1$ be parallel to the vector $-3\hat{i} + 2\hat{j} + 4\hat{k}$ and pass through the point $(2, 6, 7)$, and the line $L_2$ be parallel to the vector $2\hat{i} + \hat{j} + 3\hat{k}$ and pass through the point $(4, 3, 5)$. If the line $L_3$ is parallel to the vector $-3\hat{i} + 5\hat{j} + 16\hat{k}$ and intersects the lines $L_1$ and $L_2$ at the points $C$ and $D$, respectively, then $\left|\overrightarrow{CD}\right|^2$ is equal to: