The volume of the tetrahedron with hat{i}-lam | vedclass

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(C) The volume of a tetrahedron with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by $V = \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]|$.Given edg

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$7$

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(C) The volume of a tetrahedron with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by $V = \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]|$.Given edges are $\vec{a} = \hat{i}-\lambda \hat{j}+\hat{k}$,$\vec{b} = \lambda \hat{i}-\hat{j}-\hat{k}$,and $\vec{c} = \hat{i}+\hat{j}+\lambda \hat{k}$.The volume is $2$,so $\frac{1}{6} |\vec{a} \cdot (\vec{b} 	imes \vec{c})| = 2$,which implies $|\vec{a} \cdot (\vec{b} 	imes \vec{c})| = 12$.The scalar triple product is given by the determinant:$|\det \begin{bmatrix} 1 & -\lambda & 1 \ \lambda & -1 & -1 \ 1 & 1 & \lambda \end{bmatrix}| = 12$.Expanding the determinant:$1(-\lambda - (-1)) -(-\lambda)(\lambda^2 - (-1)) + 1(\lambda - (-1)) = \pm 12$.$(1-\lambda) + \lambda(\lambda^2+1) + (\lambda+1) = \pm 12$.$2 + \lambda^3 + \lambda = \pm 12$.Case $1$: $\lambda^3 + \lambda + 2 = 12 \Rightarrow \lambda^3 + \lambda - 10 = 0$. For $\lambda = 2$,$8 + 2 - 10 = 0$. So $\lambda = 2$ is a solution.Case $2$: $\lambda^3 + \lambda + 2 = -12 \Rightarrow \lambda^3 + \lambda + 14 = 0$. No integer solution exists for this equation.Thus,$\lambda = 2$.We need to find $|\lambda \hat{i} - 3\lambda \hat{j} + 3\hat{k}| = |2\hat{i} - 6\hat{j} + 3\hat{k}|$.Magnitude $= \sqrt{2^2 + (-6)^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7$.

The volume of the tetrahedron with $\hat{i}-\lambda \hat{j}+\hat{k}$,$\lambda \hat{i}-\hat{j}-\hat{k}$ and $\hat{i}+\hat{j}+\lambda \hat{k}$ as coterminous edges is $2$. If $\lambda$ is an integer,then $|\lambda \hat{i}-3 \lambda \hat{j}+3 \hat{k}|=$

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