The correct order of steps from A to E, for finding value of p, so that the lines $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$ and $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$ are at right angle is:

A. $p = \frac{70}{11}$

B. $(-3) \times \left(\frac{-3p}{7}\right) + 1 \times \left(\frac{2p}{7}\right) + 2 \times (-5) = 0$

C. $\frac{x-1}{-3} = \frac{y-2}{2\frac{p}{7}} = \frac{z-3}{2}$, $\frac{x-1}{-\frac{3p}{7}} = \frac{y-5}{1} = \frac{z-6}{-5}$

D. $\frac{9p}{7} + \frac{2p}{7} - 10 = 0$

E. $\frac{11p}{7} = 10$

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The value of $\lambda$, so that the lines $\frac{1-x}{3} = \frac{7y-14}{2\lambda} = \frac{z-3}{2}$ and $\frac{7-7x}{3\lambda} = \frac{y-5}{1} = \frac{6-z}{5}$ are perpendicular is:

The Cartesian equation of the line passing through the point (1, 2, -1) and parallel to the line $5x - 25 = 14 - 7y = 35z$ is

The distance of the plane $\vec{r} \cdot \left(\frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}\right) = 2$ from the origin is

If the lines $\frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{-1}$ and $x = \frac{y + 1/2}{2\lambda} = \frac{z - 1}{3}$ are perpendicular, then the value of $\lambda$ is equal to

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

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The shortest distances of the point $(1, 2, 3)$ from $x$, $y$, $z$ axes respectively are