The direction cosines of the projection of the line $\frac{x}{- 2} = \frac{y - 1}{1} = \frac{z + 1}{1}$ on the plane $2 x + y - 3 z = 5$

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$(2, - 1, 1)$

$(\frac{2}{7}, \frac{- 1}{7}, \frac{1}{7})$

$(\frac{- 2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{- 1}{\sqrt{6}})$

$(\frac{2}{\sqrt{6}}, \frac{- 1}{\sqrt{6}}, \frac{1}{\sqrt{6}})$

answer is D.

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Detailed Solution

$L e t P (α, β, γ), t h e n 2 α + β - 3 γ = 5_____(1) a n d \frac{α - 0}{2} = \frac{β - 1}{1} = \frac{γ + 1}{- 3}_____(2)$

From 1 and 2 $(α, β, γ) = (\frac{1}{7}, \frac{15}{14}, \frac{- 17}{14})$
Let $Q = (α', β', γ')$ then $2 α' + β - 3 γ' = 5____(3)$ and $\frac{α' + 2}{2} = \frac{β' - 2}{1} = \frac{γ' + 2}{- 3}____(4)$
Solving (3) and (4); $α' = \frac{- 13}{7} | | β' = \frac{29}{14} | | γ' = \frac{- 31}{14}$
d.r’s of PQ are $(\frac{1}{7}) (\frac{- 13}{7}) + (\frac{15}{14}) (\frac{29}{14}) + (\frac{- 17}{14}) (\frac{- 31}{14})$
i.e. $2; - 1; 1$
d.is $(\frac{2}{\sqrt{6}}, \frac{- 1}{\sqrt{6}}, \frac{1}{\sqrt{6}})$