A) 1
B) 2
C) 3
D) 0
Correct Answer: D
Solution :
The given lines are coplanar, \[\left| \begin{matrix} 3-1 & 1-3 & 3-1 \\ 1 & 2 & -\lambda \\ \lambda & 3 & 4 \\ \end{matrix} \right|=0=\left| \begin{matrix} 2 & -2 & 2 \\ 1 & 2 & -\lambda \\ \lambda & 3 & 4 \\ \end{matrix} \right|\] Applying \[{{C}_{2}}\to {{C}_{2}}+{{C}_{1}},\] \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\] \[\Rightarrow \] \[\left| \begin{matrix} 2 & 0 & 0 \\ 1 & 3 & -\lambda -1 \\ \lambda & \lambda +3 & 4-\lambda \\ \end{matrix} \right|\] \[\Rightarrow \] \[{{\lambda }^{2}}+\lambda +15=0\] Which has normal value of \[\lambda \] \[\therefore \] \[D=1-4\times 15\] \[=-59<0\]
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