A) 0
B) -1
C) 2
D) 1
Correct Answer: A
Solution :
Since \[\overset{\to }{\mathop{a}}\,=\hat{i}-2\hat{j}+3\hat{k},\overset{\to }{\mathop{b}}\,=2\hat{i}+3j-\hat{k}\]and \[\overset{\to }{\mathop{c}}\,=\lambda \hat{i}+\hat{j}+\left( 2\lambda -1 \right)\hat{k}\]are coplanar therefore \[[\vec{a}\,\vec{b}\,\vec{c}]=0\]i.e.,\[\left| \begin{matrix} 1 & 2 & \lambda \\ -2 & 3 &1 \\ 3 & -1 & 2\lambda -1 \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[1(6\lambda -2)-2(-4\lambda -1)+\lambda (-7)=0\] \[\Rightarrow \]\[(6\lambda -2)-8\lambda +2+2+2\lambda -9\lambda =0\] \[\Rightarrow \]\[7\lambda =0\Rightarrow \lambda =0\]You need to login to perform this action.
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