a \cdot (\vec b \times \vec c)\right| $$

Given volume is $11$ cubic units, so

$$ \frac{1}{6}\left|\vec a \cdot (\vec b \times \vec c)\right|=11 $$

$$ \left|\vec a \cdot (\vec b \times \vec c)\right|=66 $$

Now,

$$ \vec a=-2\hat i+3\hat j-3\hat k $$

$$ \vec b=4\hat i+5\hat j+(\lambda-10)\hat k $$

$$ \vec c=6\hat i+2\hat j-3\hat k $$

The scalar triple product is

$$ \vec a \cdot (\vec b \times \vec c) = \begin{vmatrix} -2 & 3 & -3 \\ 4 & 5 & \lambda-10 \\ 6 & 2 & -3 \end{vmatrix} $$

Expanding,

$$ = -2 \begin{vmatrix} 5 & \lambda-10 \\ 2 & -3 \end{vmatrix} -3 \begin{vmatrix} 4 & \lambda-10 \\ 6 & -3 \end{vmatrix} -3 \begin{vmatrix} 4 & 5 \\ 6 & 2 \end{vmatrix} $$

$$ = -2[-15-2(\lambda-10)]-3[-12-6(\lambda-10)]-3[8-30] $$

$$ = -2(5-2\lambda)-3(48-6\lambda)+66 $$

$$ = -10+4\lambda-144+18\lambda+66 $$

$$ =22\lambda-88 $$

So,

$$ |22\lambda-88|=66 $$

$$ 22|\lambda-4|=66 $$

$$ |\lambda-4|=3 $$

Hence,

$$ \lambda-4=3 \quad \text{or} \quad \lambda-4=-3 $$

So,

$$ \lambda=7 \quad \text{or} \quad \lambda=1 $$

Sum of possible values:

$$ 7+1=8 $$

So the correct option is Option B: $8$.

Vector Algebra

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