Let bar{a}=lambda bar{i}+3 bar{j}+4 bar{k}, b | vedclass

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(D) The volume of a parallelepiped with coterminus edges $\bar{a}, \bar{b}, \bar{c}$ is given by the absolute value of the scalar triple product $|[\b

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$1$

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(D) The volume of a parallelepiped with coterminus edges $\bar{a}, \bar{b}, \bar{c}$ is given by the absolute value of the scalar triple product $|[\bar{a} \bar{b} \bar{c}]|$.Given $|[\bar{a} \bar{b} \bar{c}]| = 61$.$\begin{vmatrix} \lambda & 3 & 4 \ 3 & -1 & \lambda \ \lambda & 1 & -3 \end{vmatrix} = \pm 61$.Expanding the determinant along the first row:$\lambda(3 - \lambda) - 3(-9 - \lambda^2) + 4(3 + \lambda) = \pm 61$.$3\lambda - \lambda^2 + 27 + 3\lambda^2 + 12 + 4\lambda = \pm 61$.$2\lambda^2 + 7\lambda + 39 = \pm 61$.Case $1$: $2\lambda^2 + 7\lambda + 39 = 61 \Rightarrow 2\lambda^2 + 7\lambda - 22 = 0$.Using the quadratic formula: $\lambda = \frac{-7 \pm \sqrt{49 - 4(2)(-22)}}{2(2)} = \frac{-7 \pm \sqrt{49 + 176}}{4} = \frac{-7 \pm \sqrt{225}}{4} = \frac{-7 \pm 15}{4}$.$\lambda = \frac{8}{4} = 2$ or $\lambda = \frac{-22}{4} = -5.5$.Since $\lambda$ must be an integer, $\lambda = 2$ is a solution.Case $2$: $2\lambda^2 + 7\lambda + 39 = -61 \Rightarrow 2\lambda^2 + 7\lambda + 100 = 0$.The discriminant $D = 7^2 - 4(2)(100) = 49 - 800 = -751 Thus, the only possible integer value for $\lambda$ is $2$. The number of possible values is $1$.

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