JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Matrices

From equations (1) and (3): $x+y=6-z$ and $x+2y=14-3z$. Solving gives $x=z-2$ and $y=8-2z$. Substituting into equation (2): $2(z-2)+5(8-2z)+\alpha z=\beta$, which simplifies to $(\alpha -8)z=\beta -36$. For infinitely many solutions, both coefficients must be zero: $\alpha -8=0$ and $\beta -36=0$. Thus $\alpha =8$, $\beta =36$, and $\alpha +\beta =44$.