Question

If the volume of the tetrahedron whose coterminous edges are given by the vectors \(\overset{̄}{a} = -2\hat{i}+3\hat{j}-3\hat{k}\), \(\overset{̄}{b} = 4\hat{i}+5\hat{j}+(λ-10)\hat{k}\), \(\overset{̄}{c} = 6\hat{i}+2\hat{j}-3\hat{k}\) is 11 cubic units, then the sum of the possible values of \(λ\) is \(\ldots\)

• A. \(7\)
• B. \(8\)
• C. \(1\)
• D. \(6\)

Hint:

Volume = |scalar triple product| / 6. Evaluate the determinant in terms of lambda.

Answer 1

The Correct Option is B

Step 1: Simplify the determinant using a row operation:
The determinant is linear in $\lambda$ because $\lambda$ appears in only one entry. So two test values are enough to find it.
At $\lambda = 10$ the middle row is $(4,5,0)$. Determinant: $-2(-15 - 0) - 3(-12 - 0) + (-3)(8 - 30) = 30 + 36 + 66 = 132$.
At $\lambda = 11$ the middle row is $(4,5,1)$. Determinant: $-2(-15 - 2) - 3(-12 - 6) - 3(-22) = 34 + 54 + 66 = 154$.

Step 2: Form the linear function:
The slope is $154 - 132 = 22$ per unit, so $D(\lambda) = 132 + 22(\lambda - 10) = 22\lambda - 88$.

Step 3: Apply the volume:
$|D| = 6\times11 = 66$ gives $\lambda = 7$ or $\lambda = 1$. The sum is 8.

Final Answer:
Option (B). \[ \boxed{8 \text{ (B)}} \]