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If $2 \vec{i}-\hat{j}+3 \hat{k},-12 \hat{i}-\hat{j}-3 \hat{k},-\hat{i}+2 \hat{j}-4 \hat{k}$ and $\lambda \hat{i}+2 \hat{j}-\hat{k}$ are the position vectors of four coplanar points, then $\lambda=$
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Verified Answer
The correct answer is:
$6$
$\mathrm{A}: 2 \hat{i}-\hat{j}+3 \hat{k} ; \mathrm{B}:-12 \hat{i}-\hat{j}-3 \hat{k}$
$\begin{aligned}
& \mathrm{C}:-\hat{i}+2 \hat{j}-4 \hat{k} ; \mathrm{D}: \lambda \hat{i}+2 \hat{j}-\hat{k} \\
& \overrightarrow{\mathrm{AB}}=-14 \hat{i}-6 \hat{k} ; \overrightarrow{\mathrm{BC}}=11 \hat{i}+3 \hat{j}-\hat{k} \\
& \overrightarrow{\mathrm{AC}}=-3 \hat{i}+3 \hat{j}-7 \hat{k} ; \overrightarrow{\mathrm{AD}}=(\lambda-2) \hat{i}+3 \hat{j}-4 \hat{k}
\end{aligned}$
for coplanar vectors,
$\begin{aligned}
& \overrightarrow{\mathrm{AB}} \cdot(\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{AD}})=0 \\
& \left|\begin{array}{ccc}
-14 & 0 & -6 \\
-3 & 3 & -7 \\
\lambda-2 & 3 & -4
\end{array}\right|=0 \\
& \Rightarrow 9(-14+2(3+\lambda-2))=0 \therefore \lambda=6
\end{aligned}$
$\begin{aligned}
& \mathrm{C}:-\hat{i}+2 \hat{j}-4 \hat{k} ; \mathrm{D}: \lambda \hat{i}+2 \hat{j}-\hat{k} \\
& \overrightarrow{\mathrm{AB}}=-14 \hat{i}-6 \hat{k} ; \overrightarrow{\mathrm{BC}}=11 \hat{i}+3 \hat{j}-\hat{k} \\
& \overrightarrow{\mathrm{AC}}=-3 \hat{i}+3 \hat{j}-7 \hat{k} ; \overrightarrow{\mathrm{AD}}=(\lambda-2) \hat{i}+3 \hat{j}-4 \hat{k}
\end{aligned}$
for coplanar vectors,
$\begin{aligned}
& \overrightarrow{\mathrm{AB}} \cdot(\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{AD}})=0 \\
& \left|\begin{array}{ccc}
-14 & 0 & -6 \\
-3 & 3 & -7 \\
\lambda-2 & 3 & -4
\end{array}\right|=0 \\
& \Rightarrow 9(-14+2(3+\lambda-2))=0 \therefore \lambda=6
\end{aligned}$
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