If $\mathbf{t}$ is real and $\lambda=\frac{t^2-3 t+4}{t^2+3 t+4}$ then the number of solutions of sy

Question

If $\mathbf{t}$ is real and $\lambda=\frac{t^2-3 t+4}{t^2+3 t+4}$ then the number of solutions of system of equations $3 x-y+4 z=3, x+2 y-3 z=-2$ and $6 x+5 y+\lambda z=-3$ is

Professor PC Thomas Classes · Answer

$ \begin{aligned} & f( \pm 2)=\frac{4-6+4}{4+6+4}=\frac{1}{7} \Rightarrow x \in\left(\frac{1}{7}, 7ight) \ & \left|\begin{array}{ccc} 3 & -1 & 4 \ 1 & 2 & -3 \ 6 & 5 & \lambda \end{array}ight| \end{aligned} $ $$ =3(2 \lambda+15)+1(\lambda+18)+4(-7)=7 \lambda+35 $$ For $\lambda \in\left(\frac{1}{7}, 7ight), 7 \lambda+35 
eq 0 \Rightarrow$ the system has unique solution