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(B) The volume of a tetrahedron with coterminous edges $\vec{a}$, $\vec{b}$, and $\vec{c}$ is given by $V = \frac{1}{6} |\vec{a} \cdot (\vec{b} 	imes

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$8$

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(B) The volume of a tetrahedron with coterminous edges $\vec{a}$, $\vec{b}$, and $\vec{c}$ is given by $V = \frac{1}{6} |\vec{a} \cdot (\vec{b} 	imes \vec{c})|$.Given $V = 11$, we have $|\vec{a} \cdot (\vec{b} 	imes \vec{c})| = 66$.The scalar triple product is the determinant of the matrix formed by the components:$\vec{a} \cdot (\vec{b} 	imes \vec{c}) = \begin{vmatrix} -2 & 3 & -3 \ 4 & 5 & \lambda - 10 \ 6 & 2 & -3 \end{vmatrix}$.Expanding along the first row:$= -2[5(-3) - 2(\lambda - 10)] - 3[4(-3) - 6(\lambda - 10)] - 3[4(2) - 6(5)]$$= -2[-15 - 2\lambda + 20] - 3[-12 - 6\lambda + 60] - 3[8 - 30]$$= -2[5 - 2\lambda] - 3[48 - 6\lambda] - 3[-22]$$= -10 + 4\lambda - 144 + 18\lambda + 66 = 22\lambda - 88$.Setting $|22\lambda - 88| = 66$, we get $22\lambda - 88 = 66$ or $22\lambda - 88 = -66$.Case $1$: $22\lambda = 154 \implies \lambda = 7$.Case $2$: $22\lambda = 22 \implies \lambda = 1$.The sum of possible values of $\lambda$ is $7 + 1 = 8$.

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