Let vec{a} = hat{i} + 2hat{j} + 4hat{k},vec{b | vedclass

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(D) Since $\vec{a}, \vec{b}, \vec{c}$ are coplanar,their scalar triple product is zero: $[\vec{a} \vec{b} \vec{c}] = 0$.$\begin{vmatrix} 1 & 2 & 4 \

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$-10\hat{i} + 5\hat{j}$

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(D) Since $\vec{a}, \vec{b}, \vec{c}$ are coplanar,their scalar triple product is zero: $[\vec{a} \vec{b} \vec{c}] = 0$.$\begin{vmatrix} 1 & 2 & 4 \ 1 & \lambda & 4 \ 2 & 4 & \lambda^2 - 1 \end{vmatrix} = 0$.Applying $R_3 ightarrow R_3 - 2R_1$:$\begin{vmatrix} 1 & 2 & 4 \ 1 & \lambda & 4 \ 0 & 0 & \lambda^2 - 9 \end{vmatrix} = 0$.Expanding along $R_3$:$(\lambda^2 - 9)(\lambda - 2) = 0$.This gives $\lambda = 2$ or $\lambda^2 = 9$. If $\lambda = 2$,then $\vec{a}$ and $\vec{b}$ are identical,which is trivial. For non-zero $\vec{a} 	imes \vec{c}$,we consider $\lambda^2 = 9$ (i.e.,$\lambda = 3$ or $\lambda = -3$).$\vec{a} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 4 \ 2 & 4 & \lambda^2 - 1 \end{vmatrix} = \hat{i}(2\lambda^2 - 2 - 16) - \hat{j}(\lambda^2 - 1 - 8) + \hat{k}(4 - 4) = (2\lambda^2 - 18)\hat{i} - (\lambda^2 - 9)\hat{j}$.Substituting $\lambda^2 = 9$:$\vec{a} 	imes \vec{c} = (2(9) - 18)\hat{i} - (9 - 9)\hat{j} = 0\hat{i} - 0\hat{j} = \vec{0}$.Re-evaluating the determinant: The condition for coplanarity is $(\lambda^2 - 9)(\lambda - 2) = 0$. If $\lambda = 2$,$\vec{a} 	imes \vec{c} = (2(4) - 18)\hat{i} - (4 - 9)\hat{j} = -10\hat{i} + 5\hat{j}$.

Let $\vec{a} = \hat{i} + 2\hat{j} + 4\hat{k}$,$\vec{b} = \hat{i} + \lambda\hat{j} + 4\hat{k}$,and $\vec{c} = 2\hat{i} + 4\hat{j} + (\lambda^2 - 1)\hat{k}$ be coplanar vectors. Then the non-zero vector $\vec{a} \times \vec{c}$ is:

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