Given that system of linear equations
\(\begin{aligned}
x+y+z & =a \quad \ldots (i)\\
x-y+b z & =2 \quad \ldots (ii) \\
2 x+3 y-z & =1 \quad \ldots (iii)
\end{aligned}\)
has infinitely many solutions,
\(\begin{aligned}
& \therefore \quad\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & b \\
2 & 3 & -1
\end{array}\right]=0 \\
& \Rightarrow \quad 1(1-3 b)-1(-1-2 b)+1(3+2=0 \\
& \Rightarrow \quad 1-3 b+1+2 b+5=0 \\
& \Rightarrow \\
& -b=-7 \\
& \Rightarrow \\
& b=7
\end{aligned}\)
Adding Eqs. (i) and (ii), we get
\(\quad \begin{aligned}
2 x+8 z & =a+2 \\
\Rightarrow x+4 z & =\frac{a+2}{2} \quad \ldots (iv)
\end{aligned}\)
Multiply Eq. (i) by 3 then subtract Eq. (iii) from it,
\(\begin{aligned}
3 x+3 y+3 z & =3 a \\
-2 x \pm 3 y \mp z & =-1 \\
\hline x+4 z & =3 a-1 \quad \ldots (v)
\end{aligned}\)
By Eqs. (iv) and (v), we get
\(\begin{array}{rlrl}
& & \frac{a+2}{2} & =3 a-1 \\
\Rightarrow & & a+2 & =6 a-2 \\
\Rightarrow & 6 a-a & =4 \\
\Rightarrow & & 5 a & =4 \\
\therefore & b-5 a & =7-4=3
\end{array}\)