A plane P is parallel to two lines whose direction ratios are $- 2, 1, - 3, a n d - 1, 2, - 2$ and it contains the point $(2, 2, - 2) .$ Let P intersect the co-ordnate axes at the points A, B, C making the intercepts $α, β, γ$ . If V is the volume of the tetrahedron OABC, where O is the origin and $p = α + β + γ,$ then the ordered pair (V, p) is equal to

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$(48, - 13)$

$(24, - 13)$

$(48, 11)$

$(24, - 5)$

answer is B.

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Detailed Solution

$N o r m a l o f p l a n e P = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 2 & 1 & - 3 \\ - 1 & 2 & - 2 \end{matrix}| = 4 \hat{i} - \hat{j} - 3 \hat{k} E q u a t i o n o f p l a n e P w h i c h p a s s e s t h r o u g h (2, 2, - 2) i s 4 (x - 2) - (y - 2) - 3 (z + 2) = 0 4 x - y - 3 z - 12 = 0 The plane intersects coordinate axes at A (3, 0, 0), B (0, - 12, 0), C (0, 0, - 4) \Rightarrow α = 3, β = - 12, γ = - 4 \Rightarrow p = α + β + γ = - 13 Now, volume of tetrahedron OABC is V = |\frac{1}{6} \overset{⇀}{O A} . (\overset{⇀}{O B} X \overset{⇀}{O C})| = \frac{1}{6} |\begin{matrix} 3 & 0 & 0 \\ 0 & - 12 & 0 \\ 0 & 0 & - 4 \end{matrix}| = 24 (V . p) = (24, - 13)$