• # question_answer The lines $\frac{x-3}{1}=\frac{y-}{2}=\frac{z-3}{-\lambda }$ and $\frac{x-1}{\lambda }=\frac{y-3}{3}=\frac{z-1}{4}$ are coplanar. Then, the number of possible value(s) of $\lambda$ is A)  1                                 B)  2 C)  3                                 D)  0

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 $\lambda$ $\therefore$    $D=1-4\times 15$ $=-59<0$