The shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{4} = \frac{y-4}{6} = \frac{z-5}{8}$ is equal to

If $\lambda$ and $\mu$ are parameters, then

Match the LIST-I with LIST-II

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If the lines $x = ay+b$, $z = cy+d$ and $x = py+q$, $z = ry+s$ are perpendicular, then $ap + cr$ is equal to

A line makes angle $\theta$ with $x$-axes as well as $z$-axis. If the angle $\beta$, which it makes with $y$-axis is such that $\sin^2\beta = 3\sin^2\theta$, then $\cos^2\theta$ is :

The distance of plane $\vec{r}\cdot(6\hat{i} - 3\hat{j} - 2\hat{k}) + 1 = 0$ from origin is :

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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: