For the system of linear equations

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

$x+2y+3z=4$

$2x+5y+2z=8$

which of the following is $\mathrm{NOT}$ correct?

Given,

System of linear equations,

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

$x+2y+3z=4$

$x+2y+3z=4$

Now finding,

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

$\Rightarrow △=-18\left(a-3\right)$

Hence, for $a=3, △=0$

Now finding,

${△}_3=\left|\begin{array}{ccc}b& 2& 4\\ 4& 1& 2\\ 8& 2& -5\end{array}\right|$

$\Rightarrow {△}_3=9\left(b-8\right)$

So, for $b=8, {△}_3=0$

Now solving options we get,

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

For $a=8,b=8, △\ne 0, {△}_3=0\Rightarrow$unique solution

For $a=b=6, △\ne 0, {△}_3\ne 0\Rightarrow$ unique solution

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