Let $P_{1} : 2 x + y - z = 3$ and $P_{2} : x + 2 y + z = 2$ be two planes. Then, which of the following statements(s) is (are) TRUE?

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The line $\frac{3 x - 4}{9} = \frac{1 - 3 y}{9} = \frac{z}{3}$ is perpendicular to the line of intersection of $P_{1}$ and $P_{2}$

The acute angle between $P_{1}$ and $P_{2}$ is $60 °$

If $P_{3}$ is the plane passing through the point (4,2,−2) and perpendicular to the line

of intersection of $P_{1}$ and $P_{2}$ , then the distance of the point (2,1,1) from the plane $P_{3}$ is $\frac{2}{\sqrt{3}}$

The line of intersection of $P_{1}$ and $P_{2}$ has direction ratios 1,2,−1

answer is C, D.

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Detailed Solution

The direction cosines of line of intersection of two planes is (a,b,c)

$\begin{array}{l} a + 2 b + c = 0, \\ 2 a + b - c = 0 \\ \frac{a}{1 + 2} = \frac{b}{- 1 - 2} = \frac{c}{4 - 1} \Rightarrow \frac{a}{3} = \frac{b}{- 3} = \frac{c}{3} \end{array}$

$\Rightarrow$ 3,-3,3 are direction ratios of the line of intersection of planes.

hence the option (1) is wrong

$\frac{3 x - 4}{9} = \frac{1 - 3 y}{9} = \frac{z}{3} \Rightarrow \Rightarrow \frac{x - 4 / 3}{3} = \frac{y - 1 / 3}{- 3} = \frac{z}{3}$ , lines are parallel

hence the option (2) is wrong

Acute angle between the given two planes is $= \cos^{- 1} (\frac{2 \times 1 + 1 \times 2 - 1 \times 1}{\sqrt{6} \sqrt{6}}) = = \cos^{- 1} (\frac{3}{6}) = \cos^{- 1} (\frac{1}{2}) = 60 °$