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(D) The volume $V$ of the parallelepiped formed by vectors $\vec{a} = \hat{i} + \lambda \hat{j} + \hat{k}$,$\vec{b} = \hat{j} + \lambda \hat{k}$,and $

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(D) The volume $V$ of the parallelepiped formed by vectors $\vec{a} = \hat{i} + \lambda \hat{j} + \hat{k}$,$\vec{b} = \hat{j} + \lambda \hat{k}$,and $\vec{c} = \lambda \hat{i} + \hat{k}$ is given by the absolute value of the scalar triple product:$V = |\vec{a} \cdot (\vec{b} 	imes \vec{c})| = \left| \det \begin{bmatrix} 1 & \lambda & 1 \ 0 & 1 & \lambda \ \lambda & 0 & 1 \end{bmatrix} ight|$Expanding the determinant along the first row:$\det = 1(1 - 0) - \lambda(0 - \lambda^2) + 1(0 - \lambda) = 1 + \lambda^3 - \lambda$So,$V(\lambda) = |\lambda^3 - \lambda + 1|$.To find the minimum,let $f(\lambda) = \lambda^3 - \lambda + 1$. We find the critical points by setting $f'(\lambda) = 0$:$f'(\lambda) = 3\lambda^2 - 1 = 0 \implies \lambda^2 = \frac{1}{3} \implies \lambda = \pm \frac{1}{\sqrt{3}}$.Evaluating $f(\lambda)$ at these points:For $\lambda = \frac{1}{\sqrt{3}}$,$f\left(\frac{1}{\sqrt{3}}ight) = \frac{1}{3\sqrt{3}} - \frac{1}{\sqrt{3}} + 1 = 1 - \frac{2}{3\sqrt{3}} \approx 1 - 0.385 = 0.615$.For $\lambda = -\frac{1}{\sqrt{3}}$,$f\left(-\frac{1}{\sqrt{3}}ight) = -\frac{1}{3\sqrt{3}} + \frac{1}{\sqrt{3}} + 1 = 1 + \frac{2}{3\sqrt{3}} \approx 1.385$.Since the volume $V = |f(\lambda)|$,we look for the minimum of $|f(\lambda)|$. The function $f(\lambda)$ has a local minimum at $\lambda = \frac{1}{\sqrt{3}}$ with value $\approx 0.615$. Since $f(\lambda) 	o -\infty$ as $\lambda 	o -\infty$ and $f(\lambda) 	o \infty$ as $\lambda 	o \infty$,the function $f(\lambda)$ crosses zero at some $\lambda Since none of the provided options correspond to the root of $\lambda^3 - \lambda + 1 = 0$,the correct option is $D$.

If the volume of the parallelepiped formed by the vectors $\hat{i} + \lambda \hat{j} + \hat{k}$,$\hat{j} + \lambda \hat{k}$ and $\lambda \hat{i} + \hat{k}$ is minimum,then $\lambda$ is equal to

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