For the system of linear equations 2x+4y+2az=b x+2y+3z=4 2x+5y+2z=8 which of the following is NOT correct?

Given,

System of linear equations,

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

$$x+2y+3z=4$$x+2y+3z=4

$$x+2y+3z=4$$x+2y+3z=4

Now finding,

$$\triangle =|\begin{matrix}2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & -5 & 2\end{matrix}|$$△=∣212​42−5​2a32​∣

$$\Rightarrow \triangle =-18(a-3)$$⇒△=−18(a−3)

Hence, for $a=3,\triangle =0$a=3,△=0

Now finding,

$${\triangle }_{3}=|\begin{matrix}b & 2 & 4 \\ 4 & 1 & 2 \\ 8 & 2 & -5\end{matrix}|$$△3​=∣b48​212​42−5​∣

$$\Rightarrow {\triangle }_{3}=9(b-8)$$⇒△3​=9(b−8)

So, for $b=8,{\triangle }_{3}=0$b=8,△3​=0

Now solving options we get,

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

For $a=8,b=8,\triangle \neq 0,{\triangle }_{3}=0\Rightarrow$a=8,b=8,△=0,△3​=0⇒unique solution

For $a=b=6,\triangle \neq 0,{\triangle }_{3}\neq 0\Rightarrow$a=b=6,△=0,△3​=0⇒ unique solution

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