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

Given,

System of linear equations,

$2x + 4y + 2az = b$

$x + 2y + 3z = 4$

$x + 2y + 3z = 4$

Now finding,

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

$⇒ △ = - 18\left(a - 3\right)$

Hence, for $a = 3, △ = 0$

Now finding,

{△}_{3} = \left|\begin{cases}b & 2 & 4 \\ 4 & 1 & 2 \\ 8 & 2 & - 5\end{cases}\right|

$⇒ {△}_{3} = 9\left(b - 8\right)$

So, for $b = 8, {△}_{3} = 0$

Now solving options we get,

If $a = 3$ and $b = 8$ we'll have infinite solution

For $a = 8, b = 8, △ \ne 0, {△}_{3} = 0 ⇒$unique solution

For $a = b = 6, △ \ne 0, {△}_{3} \ne 0 ⇒$ unique solution

$∴$ For $a = 3, b = 6$ we'll have no solution