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(D) The volume $V$ of the parallelepiped formed by vectors $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $|[\vec{a} \vec{b} \vec{c

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$\frac{1}{\sqrt{3}}$

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(D) The volume $V$ of the parallelepiped formed by vectors $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $|[\vec{a} \vec{b} \vec{c}]|$.$V = \begin{vmatrix} 1 & m & 1 \ 0 & 1 & m \ m & 0 & 1 \end{vmatrix} = 1(1 - 0) - m(0 - m^2) + 1(0 - m) = 1 + m^3 - m$.To find the minimum volume,we differentiate $V$ with respect to $m$:$\frac{dV}{dm} = 3m^2 - 1$.Setting $\frac{dV}{dm} = 0$,we get $3m^2 = 1$,which implies $m^2 = \frac{1}{3}$,so $m = \pm \frac{1}{\sqrt{3}}$.Since we are looking for the volume (which must be positive),we consider the magnitude. For $m > 0$,we take $m = \frac{1}{\sqrt{3}}$.Checking the second derivative: $\frac{d^2V}{dm^2} = 6m$.For $m = \frac{1}{\sqrt{3}}$,$\frac{d^2V}{dm^2} = 6(\frac{1}{\sqrt{3}}) > 0$,which confirms that the volume is minimum at $m = \frac{1}{\sqrt{3}}$.

The volume of the parallelepiped formed by the vectors $\hat{i} + m \hat{j} + \hat{k}$,$\hat{j} + m \hat{k}$,and $m \hat{i} + \hat{k}$ becomes minimum when $m$ is

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