If the volume of the tetrahedron whose vertices are (1,−6,10),(1,−3,7)(5,−1,λ) and (7,−4,7), is 11 cubic units, then the sum of the values of λ, is

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If the volume of the tetrahedron whose vertices are $(1, - 6, 10), (1, - 3, 7) (5, - 1, λ) and (7, - 4, 7)$ , is $11$ cubic units, then the sum of the values of $λ$ , is

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answer is 8.

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Detailed Solution

The vertices are $A (1, - 6, 10), B (- 1, - 3, 7)$ , $C (5, - 1, λ) and D (7, - 4, 7)$

$\begin{array}{l} ∴ \vec{A B} = - 2 \hat{i} + 3 \hat{j} - 3 \hat{k}, \vec{A C} = 4 \hat{i} + 5 \hat{j} + (λ - 10) \hat{k} \\ \vec{A D} = 6 \hat{i} + 2 \hat{j} - 3 \hat{k} \end{array}$

It is given that the volume of the tetrahedron $A B C D$ is $11$ cubic units.

$\begin{array}{l} \frac{1}{6} [\begin{matrix} \vec{A B} & \vec{A C} & \vec{A D} \end{matrix}] = \pm 11 \\ \Rightarrow |\begin{array}{ccc} - 2 & 3 & - 3 \\ 4 & 5 & λ - 10 \\ 6 & 2 & - 3 \end{array}| = \pm 66 \end{array}$

$\Rightarrow |\begin{array}{ccc} - 2 & 3 & - 3 \\ 0 & 0 & λ - 4 \\ 6 & 2 & - 3 \end{array}| = \pm 66$ Applying $R_{2} \to R_{2} - (R_{1} + R_{3})$

$\begin{array}{l} \Rightarrow - (λ - 4) (- 4 - 18) = \pm 66 \\ \Rightarrow 22 (λ - 4) = \pm 66 \\ \Rightarrow λ - 4 = \pm 3 \Rightarrow λ = 7, 1 \end{array}$