Let the system of linear equations x + y + az = 2 3x + y + z = 4 x...

Given,
$x + y + az = 2$

$3x + y + z = 4$

$x + 2z = 1$
Now finding,

\Delta = \left|\begin{cases}1 & 1 & \alpha \\ 3 & 1 & 1 \\ 1 & 0 & 2\end{cases}\right| = - \left(\alpha + 3\right)

{\Delta}_{1} = \left|\begin{cases}2 & 1 & \alpha \\ 4 & 1 & 1 \\ 1 & 0 & 2\end{cases}\right| = - \left(3 + \alpha\right)

{\Delta}_{2} = \left|\begin{cases}1 & 2 & \alpha \\ 3 & 4 & 1 \\ 1 & 1 & 2\end{cases}\right| = - \left(\alpha + 3\right)

{\Delta}_{3} = \left|\begin{cases}1 & 1 & 2 \\ 3 & 1 & 4 \\ 1 & 0 & 1\end{cases}\right| = 0

$\alpha \ne - 3, x = \dfrac{{△}_{1}}{△} = 1, y = \dfrac{{△}_{2}}{△} = 1, z = \dfrac{{△}_{3}}{△} = 0,$

Now points \left(\alpha, 1\right), \left(1, \alpha\right) \text{ and } \left(1, - 1\right)are collinear</p><p>\left|\begin{cases}\alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & - 1 & 1\end{cases}\right| = 0

$⇒ \alpha\left(\alpha + 1\right) - 1\left(1 - 1\right) + 1\left(- 1 - \alpha\right) = 0$

${\alpha}^{2} + \alpha - 1 - \alpha = 0$

$\alpha = \pm 1$

So, \left|1\right| + \left|- 1\right| = 2$