A) 2
B) \[-1\]
C) 1
D) \[-2\]
Correct Answer: D
Solution :
Given that the vectors\[\overrightarrow{a}+\lambda \overrightarrow{b}+3\overrightarrow{c},\]\[-2\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c}\]and\[\overrightarrow{a}-3\overrightarrow{b}+5\overrightarrow{c}\] are coplanar, then \[\left| \begin{matrix} 1 & \lambda & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \\ \end{matrix} \right|=0\] Applying \[{{R}_{2}}\to {{R}_{2}}+2{{R}_{1}},{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\] \[\left| \begin{matrix} 1 & \lambda & 3 \\ 0 & 3+2\lambda & 2 \\ 0 & -3-\lambda & 2 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[2(3+2\lambda )+2(3+\lambda )=0\] \[\Rightarrow \] \[6+4\lambda +6+2\lambda =0\] \[\Rightarrow \] \[6\lambda =-12\] \[\Rightarrow \] \[\lambda =-2\]
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