If 3 hat{i}+3 hat{j}+sqrt{3} hat{k}, hat{i}+h | vedclass

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(A) Let $\vec{a} = 3 \hat{i} + 3 \hat{j} + \sqrt{3} \hat{k}$, $\vec{b} = \hat{i} + \hat{k}$, and $\vec{c} = \sqrt{3} \hat{i} + \sqrt{3} \hat{j} + \lam

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(A) Let $\vec{a} = 3 \hat{i} + 3 \hat{j} + \sqrt{3} \hat{k}$, $\vec{b} = \hat{i} + \hat{k}$, and $\vec{c} = \sqrt{3} \hat{i} + \sqrt{3} \hat{j} + \lambda \hat{k}$.Since these vectors are coplanar, their scalar triple product must be zero, i.e.,$\vec{a} \cdot (\vec{b} 	imes \vec{c}) = 0$.This is equivalent to the determinant of the matrix formed by their components being zero:$\begin{vmatrix} 3 & 3 & \sqrt{3} \ 1 & 0 & 1 \ \sqrt{3} & \sqrt{3} & \lambda \end{vmatrix} = 0$Expanding the determinant along the first row:$3(0 - \sqrt{3}) - 3(\lambda - \sqrt{3}) + \sqrt{3}(\sqrt{3} - 0) = 0$$-3\sqrt{3} - 3\lambda + 3\sqrt{3} + 3 = 0$$-3\lambda + 3 = 0$$3\lambda = 3$$\lambda = 1$

If $3 \hat{i}+3 \hat{j}+\sqrt{3} \hat{k}$, $\hat{i}+\hat{k}$, and $\sqrt{3} \hat{i}+\sqrt{3} \hat{j}+\lambda \hat{k}$ are coplanar, then $\lambda$ is equal to

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