A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) \[\frac{7}{2}\]
D) \[\frac{7}{12}\]
Correct Answer: D
Solution :
\[\left| \begin{matrix} 2 & 1 & 1 \\ 3 & 2\lambda & 4 \\ 1 & 1 & -3 \\ \end{matrix} \right|=0;\] \[\left| \begin{matrix} 1 & 0 & 1 \\ 3-3\lambda & 2\lambda -4 & 4 \\ 0 & 1+3\lambda & -3\lambda \\ \end{matrix} \right|\,=0;\] \[\left| \begin{matrix} 0 & 0 & 1 \\ -2\lambda -1 & 2\lambda -4 & 4 \\ 0 & 1+3\lambda & -3\lambda \\ \end{matrix} \right|=0\] \[\Rightarrow \,(3\lambda +1)\,(2\lambda +1)\,+3\lambda (2\lambda -4)=0\] \[\Rightarrow \,6{{\lambda }^{2}}+5\lambda +1\,+6{{\lambda }^{2}}\,-12\lambda =0\] \[\Rightarrow \,12{{\lambda }^{2}}\,-7\lambda +1=0\] \[\Rightarrow \,(3\lambda -1)\,(4\lambda -1)=0\] \[\Rightarrow \,\lambda =\frac{1}{3},\,\frac{1}{4}\,\Rightarrow \,Sum\,=\frac{7}{12}\]
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