Consider the planes $P_{1}, P_{2}$ and points A(2, 5, 1) and B(1, 3, 0) where
$P_{1} : x + y + z = 3 P_{2} : 2 x + y - z = 4$
Then identify the correct statement(s).

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Shortest distance between line AB and points common to $P_{1}, P_{2}$ is equal to $\frac{1}{5 \sqrt{3}}$ units.

Projection of segment AB on $P_{1}$ is equal to $\frac{\sqrt{6}}{3}$ units.

Projection of segment AB on $P_{1}$ is equal to $2 \sqrt{2}$

Area enclosed by locus of all points on $P_{1}$ which are at a distance of 3 units from B is equal to $\frac{26 π}{3}$ sq. units.

answer is A, B, D.

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Detailed Solution

(A) $\bar{n_{1}} \times \bar{n_{2}} = < - 2, 3, - 1 >$
Point common to $p_{1}$ and $p_{2}$ is (1, 2, 0)
$∴$ Distance between skew lines is
$|\frac{(\bar{a} - \bar{c}) . (\bar{p} \times \bar{q})}{| \bar{p} \times \bar{q} |}| = \frac{1}{5 \sqrt{3}}$
(B) $\bar{B A} = \hat{i} + 2 \hat{j} + \hat{k}$ Angle between line and plane
= $\sin^{- 1} (\frac{4}{\sqrt{3} \sqrt{6}}) = θ$
Projection is $| \bar{B A} | \cos θ = \frac{\sqrt{6}}{3}$
(D) $r = \sqrt{\frac{26}{3}}$