The shortest distance between the lines $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ and $\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$ is

Two lines $L_1: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$ and $L_2: x=\alpha, \frac{y}{-1}=\frac{z}{2-\alpha}$ are coplanar. Then $\alpha$ can take value$(s)$.

If the two lines $l_{1}: \frac{x-2}{3}=\frac{y+1}{-2}, z=2$ and $l_{2}: \frac{x-1}{1}=\frac{2y+3}{\alpha}=\frac{z+5}{2}$ are perpendicular,then the angle between the lines $l_{2}$ and $l_{3}: \frac{1-x}{3}=\frac{2y-1}{-4}=\frac{z}{4}$ is

Let $l_1$ be the line passing through the point $A = 3\hat{i} + 4\hat{j} - 2\hat{k}$ and parallel to the vector $\vec{b_1} = -\hat{i} + 2\hat{j} + \hat{k}$. Let $l_2$ be another line passing through the point $B = \hat{i} - 7\hat{j} - 2\hat{k}$ and parallel to the vector $\vec{b_2} = \hat{i} + 3\hat{j} + 2\hat{k}$. Then the shortest distance between the lines $l_1$ and $l_2$ is:

If the shortest distance between the straight lines $3(x-1)=6(y-2)=2(z-1)$ and $4(x-2)=2(y-\lambda)=(z-3)$,$\lambda \in R$ is $\frac{1}{\sqrt{38}}$,then the integral value of $\lambda$ is equal to :

Consider a line $L$ passing through the points $P(1, 2, 1)$ and $Q(2, 1, -1)$. If the mirror image of the point $A(2, 2, 2)$ in the line $L$ is $(\alpha, \beta, \gamma)$,then $\alpha + \beta + 6\gamma$ is equal to: