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 :

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

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 :

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