If the system of equations $x+4 y-z=\lambda, 7 x+9 y+\mu z=-3,5 x+y+2 z=-1$ has infinitely many solutions, then $(2 \mu+3 \lambda)$ is equal to :

<p>$$\begin{aligned} & x+4 y-z=\lambda \\ & 7 x+9 y+\mu z=-3 \\ & 5 x+y+2 z=-1 \\ & {\left[\begin{array}{ccc} 1 & 4 & -1 \\ 7 & 9 & \mu \\ 5 & 1 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} \lambda \\ -3 \\ -1 \end{array}\right]} \\ & A=\left[\begin{array}{lll} 1 & 4 & -1 \\ 7 & 9 & \mu \\ 5 & 1 & 2 \end{array}\right], B=\left[\begin{array}{c} \lambda \\ -3 \\ -1 \end{array}\right] \\ & A X=B \\ & X=A^{-1} B \\ & \quad=\frac{\operatorname{adj} A}{|A|} B \end{aligned}$$</p> <p>If $|A|=0$ and $(\operatorname{adj} A) \cdot B=0$, system has infinitely many solutions.</p> <p>$$\begin{aligned} & |A|=18-\mu-4(14-5 \mu)-1(7-45)=0 \\ & \Rightarrow 18-\mu-56+20 \mu+38=0 \\ & \Rightarrow 19 \mu=0 \\ & \Rightarrow \mu=0 \\ & \text { Also adjA }=\left[\begin{array}{ccc} 18 & -9 & 9 \\ -14 & 7 & -7 \\ -38 & 19 & -19 \end{array}\right] \\ & (\text { adj } A) \cdot B=0 \\ & {\left[\begin{array}{ccc} 18 & -9 & 9 \\ -14 & 7 & -7 \\ -38 & 19 & -19 \end{array}\right]\left[\begin{array}{c} \lambda \\ -3 \\ -1 \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]} \\ & 18 \lambda+27-9=0 \\ & \Rightarrow 18 \lambda=-18 \\ & \Rightarrow \lambda=-1 \\ & \Rightarrow 2 \mu+3 \lambda=3(-1)=-3 \end{aligned}$$</p>