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If $3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\sqrt{3} \hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}, \sqrt{3} \hat{\mathbf{i}}+\sqrt{3} \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ are coplanar, then $\lambda$ is equal to
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The correct answer is:
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Let $\mathbf{a}=3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\sqrt{3} \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{k}}$ and
$\mathbf{c}=\sqrt{3} \hat{\mathbf{i}}+\sqrt{3} \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$
Since, these vectors are coplanar
$\begin{aligned}
& \therefore \quad \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=0 \\
& \Rightarrow \quad\left|\begin{array}{ccc}
3 & 3 & \sqrt{3} \\
1 & 0 & 1 \\
\sqrt{3} & \sqrt{3} & \lambda
\end{array}\right|=0 \\
& \Rightarrow \quad 3(0-\sqrt{3})-3(\lambda-\sqrt{3})+\sqrt{3}(\sqrt{3}-0)=0 \\
& \Rightarrow \quad-3 \sqrt{3}-3 \lambda+3 \sqrt{3}+3=0 \\
& \Rightarrow \quad \lambda=1
\end{aligned}$
$\mathbf{c}=\sqrt{3} \hat{\mathbf{i}}+\sqrt{3} \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$
Since, these vectors are coplanar
$\begin{aligned}
& \therefore \quad \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=0 \\
& \Rightarrow \quad\left|\begin{array}{ccc}
3 & 3 & \sqrt{3} \\
1 & 0 & 1 \\
\sqrt{3} & \sqrt{3} & \lambda
\end{array}\right|=0 \\
& \Rightarrow \quad 3(0-\sqrt{3})-3(\lambda-\sqrt{3})+\sqrt{3}(\sqrt{3}-0)=0 \\
& \Rightarrow \quad-3 \sqrt{3}-3 \lambda+3 \sqrt{3}+3=0 \\
& \Rightarrow \quad \lambda=1
\end{aligned}$
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