If the system of equations x + y + a z = b 2 x + 5 y + 2 z = 6 x + 2 y + 3 z = 3 has infinitely many…

Given system of equations are

$x+y+az=b.....\left(\mathrm{i}\right)$

$2x+5y+2z=6.....\left(\mathrm{ii}\right)$

$x+2y+3z=3.....\left(\mathrm{iii}\right)$

It is given that the system of equations have infinite solutions.

$\Rightarrow ∆=0$

$\Rightarrow \left|\begin{array}{ccc}1& 1& a\\ 2& 5& 2\\ 1& 2& 3\end{array}\right|=0$

$\Rightarrow 1\left(15-4\right)-1\left(6-2\right)+a\left(4-5\right)=0$

$\Rightarrow a=7$ 

Also, we have $\Rightarrow {∆}_1=0$

$\Rightarrow \left|\begin{array}{ccc}b& 1& a\\ 6& 5& 2\\ 3& 2& 3\end{array}\right|=0$

$\Rightarrow b\left(15-4\right)-1\left(18-6\right)+7\left(12-15\right)=0$

$\Rightarrow b=3$

$\therefore a=7, b=3$ 

$\therefore 2a+3b$$=14+9=23$

Hence, this is the correct option.