If the system of linear equations x - 2y + z = - 4 2x + + 3z = 5 3x...

Given:

The system of linear equations 

$x - 2y + z = - 4$, 

$2x + \alphay + 3z = 5$ and

$3x - y + \betaz = 3$ has infinity many solutions,

So, $△ = {△}_{1} = {△}_{2} = {△}_{3}$

Now, finding, 

{△}_{2} = \left|\begin{cases}1 & - 4 & 1 \\ 2 & 5 & 3 \\ 3 & 3 & \beta\end{cases}\right| = 0

$⇒ 5\beta - 9 + 8\beta - 36 - 9 = 0$

$⇒ 13\beta = 54$

And, {△}_{3} = \left|\begin{cases}1 & - 2 & - 4 \\ 2 & \alpha & 5 \\ 3 & - 1 & 3\end{cases}\right| = 0

$⇒ 3\alpha + 5 + 12 - 30 + 8 + 12\alpha = 0$

$⇒ 15\alpha = 5$

$⇒ 12\alpha = 4$

Hence, the value of $12\alpha + 13\beta = 4 + 54 = 58$