If the system of equations x + y + z = 6 2x + 5y + = β x + 2y + 3z...

$x + y + z = 6 . . . . \left(1\right)$

$2x + 5y + \alphaz = \beta . . . . \left(2\right)$

$x + 2y + 3z = 14 . . . \left(3\right)$

For infinite solution

\left|$\begin{cases}1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3\end{cases}$\right| = 0$</p><p>$⇒ \left(15 - 2\alpha\right) - \left(6 - \alpha\right) + \left(4 - 5\right) = 0$</p><p>$⇒ 8 - \alpha = 0 ⇒ \alpha = 8</p><p>Also \left|\begin{cases}1 & 1 & 6 \\ 2 & 5 & \beta \\ 1 & 2 & 14\end{cases}\right| = 0

$⇒ \left(70 - 2\beta\right) - \left(28 - \beta\right) + 6\left(4 - 5\right) = 0$

$⇒ 36 - \beta = 0$

i.e. $\alpha = 8, \beta = 36$

$⇒ \alpha + \beta = 44$