If the system of linear equations 7x + 11y + = 13 5x + 4y + 7z = β...

Given,

System of linear equations,

$7x + 11y + \alphaz = 13$
$5x + 4y + 7z = \beta$
$175x + 194y + 57z = 361$

Now we know that, for infinite solution $D = {D}_{1} = {D}_{2} = {D}_{3} = 0$

Now solving $D = 0$ we get,

\left|$\begin{cases}7 & 11 & \alpha \\ 5 & 4 & 7 \\ 175 & 194 & 57\end{cases}$\right| = 0$</p><p>Now applying operation${R}_{3} \to {R}_{3} - 25{R}_{1}we get,</p><p>\left|\begin{cases}7 & 11 & \alpha \\ 5 & 4 & 7 \\ 0 & - 81 & 57 - 25\alpha\end{cases}\right| = 0

$⇒ 81 ( 49 - 5\alpha ) + ( 57 - 25\alpha ) ( - 27 ) = 0$

$⇒ 270\alpha = - 2430 ⇒ \alpha = - 9$

And now solving ${D}_{1} = 0$ we get,

\left|$\begin{cases}13 & 11 & - 9 \\ \beta & 4 & 7 \\ 361 & 194 & 57\end{cases}$\right| = 0$</p><p>$⇒ 13\left(4 \times 57 - 7 \times 194\right) - \beta\left(11 \times 57 + 9 \times 194\right) + 361\left(11 \times 7 + 9 \times 4\right) = 0$</p><p>$⇒ 13\left(- 1130\right) - \beta\left(2373\right) + 361\left(113\right) = 0$</p><p>$⇒ 13\left(- 10\right) - \beta\left(21\right) + 361 = 0$</p><p>$⇒ 21\beta = 231$</p><p>$⇒ \beta = 11$</p><p>$∴ \alpha + \beta + 2 = 4$