Let $L_1$ and $L_2$ lines such that $L_1:\vec{r}=\hat{i}+\hat{j}+\lambda(2\hat{i}-\hat{j}+\hat{k})$, $L_2:\vec{r}=2\hat{i}+\hat{j}-\hat{k}+\mu(4\hat{i}-2\hat{j}+2\hat{k})$ and $\theta$ be the acute angle between $L_1$ and $L_2$, then

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	$\cos\theta$	I.	$\dfrac{\sqrt{66}}{6}$
B.	$\sin\theta$	II.	$3$
C.	the shortest distance (in units) between $L_1$ and $L_2$	III.	$1$
D.	(sum of squares of direction cosines of $L_1$) $+2$	IV.	$0$

Choose the correct answer from the options given below:

If the shortest distance between the lines $l_1$ and $l_2$ given by $\vec{r} = a\hat{i} + 2\hat{j} - \hat{k} + \lambda(2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = \hat{i} - \hat{j} + \hat{k} + \mu(2\hat{i} - \hat{j} + \hat{k})$ is $\sqrt{\frac{35}{6}}$ units, the values of 'a' can be :

The angle between the lines $\vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \lambda(\hat{i} + 2\hat{j} + 2\hat{k})$ and $\vec{r} = 5\hat{j} - 2\hat{k} + \mu(3\hat{i} + 2\hat{j} + 6\hat{k})$ is :

The image of the point (1, 6, 3) with respect to the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ is

If the line $\frac{-x+1}{3} = \frac{-y-2}{-2k} = \frac{z+3}{2}$ and $\frac{-1+x}{3k} = \frac{-1 +y}{1} = \frac{-z+6}{5}$ are perpendicular, then the value of k is:

If a line makes angles α, β and γ with positive x-axis, y-axis and z-axis respectively, then the value of

$sin^2\frac{α}{2}cos^2\frac{α}{2} + sin^2\frac{β}{2}cos^2\frac{β}{2} + sin^2\frac{γ}{2}cos^2\frac{γ}{2}$ is

The shortest distance between the following lines:

$\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + s(2\hat{i} + \hat{j} + \hat{k})$

$\vec{r} = (\hat{i} + \hat{j} + 2\hat{k}) + t(4\hat{i} + 2\hat{j} + 2\hat{k})$, where s and t are scalars, is: