JEE Main 202229 Jul, Evening shiftMathematicsMediumMatrices and Determinants245× since 2002If the system of equations x+y+z=62x+5y+αz=βx+2y+3z=14\begin{aligned} &x+y+z=6 \\ &2 x+5 y+\alpha z=\beta \\ &x+2 y+3 z=14 \end{aligned}x+y+z=62x+5y+αz=βx+2y+3z=14 has infinitely many solutions, then α+β\alpha+\betaα+β is equal toAOption A. 8BOption B. 36COption C. 44DOption D. 48CheckSkip · show solution