$\text{ If the system of equations }x+y+z=0,x+2y+3z=1,\lambda x+\mu y+4z=1\text{ has infinitely many solutions then }(\lambda ,\mu )=$

$\begin{array}{l}\mathrm{x}+\mathrm{y}+\mathrm{z}=0,\dots \dots \dots \left(1\right)\\ x+2y+3z=1\dots \dots \dots \text{ (2) }\\ \lambda x+\mu y+4z=1\dots \dots \text{ (3) }\\ 3\times \left(1\right)-\left(2\right)\text{ gives }2\mathrm{x}+\mathrm{y}=-1\dots \dots \dots ...\left(4\right)\\ 4\times \left(1\right)-\left(3\right)\text{ gives }(4-\lambda )x+(4-\mu )y=-1\dots \text{ (5) }\\ \text{ (4), (5) have infinitely many solutions }\\ \Rightarrow \frac{2}{4-\lambda }=\frac{1}{4-\mu }=1\\ \Rightarrow \lambda =2,\mu =3\end{array}$