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

$\lambda =\frac{t^2-3t+4}{t^2+3t+4}\Rightarrow \left(\lambda -1\right)t^2+3\left(\lambda +1\right)t+4\left(\lambda -1\right)=0$  Since, t is real $\Rightarrow 9{\left(\lambda +1\right)}^2-16{\left(\lambda -1\right)}^2\ge 0$  $\Rightarrow \left(3\lambda +3-4\lambda +4\right)\left(3\lambda +3+4\lambda -4\right)\ge 0$  $\Rightarrow \left(7-\lambda \right)\left(7\lambda -1\right)\ge 0\Rightarrow \frac{1}{7}\le \lambda \le 7$  Now, $\Delta =\left|\begin{array}{c}3-14\\ 12-3\\ 65\lambda \end{array}\right|$ [Determinant of coefficients of equations] $=7\left(\lambda +5\right)\ne 0\left[\therefore \frac{1}{7}\le \lambda \le 7\right]$ 

 Hence the given system of equations has a unique solution.