Given system of equations,

$$ \text { and } \quad \begin{aligned} x+y+a z & =b \\ 2 x+5 y+2 z & =6 \\ x+2 y+3 z & =3 \end{aligned} $$

Since, given system of equation has infinitely many solutions

$$ \therefore D=0 \text { and } D_1=D_2=D_3=0 $$

$$ \begin{aligned} & \text { Here, } D=\left|\begin{array}{lll} 1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3 \end{array}\right|=0 \\\\ & \Rightarrow 1(15-4)-1(6-2)+a(4-5)=0 \\\\ & \Rightarrow -a+11-4=0 \\\\ & \Rightarrow a=7 \end{aligned} $$

$$ \begin{aligned} & \text { and } D_1=\left|\begin{array}{ccc} b & 1 & a \\ 6 & 5 & 2 \\ 3 & 2 & 3 \end{array}\right|=0 \\\\ & \Rightarrow b(15-4)-1(18-6)+a(12-15)=0 \\\\ & \Rightarrow 11 b-12-3 a=0 \\\\ & \Rightarrow 11 b-12-21=0 ~~~~~~~(\because a=7)\\\\ & \Rightarrow 11 b-33=0 \\\\ & \Rightarrow b=3 \\\\ & \therefore 2 a+3 b=2(7)+3(3)=14+9=23 \end{aligned} $$