Given system of equations can written in matrix form as $AX=B$ where

$A = \left( {\begin{array}{*{20}{c}}
1&2&3\\
1&3&5\\
2&5&a
\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}
6\\
9\\
b
\end{array}} \right)$

Since, system is consistent and has infonitely many solutions

$\therefore $ (adj.$A$) $B=0$

$ \Rightarrow \left( {\begin{array}{*{20}{c}}
{3a – 25}&{15 – 2a}&1\\
{10 – a}&{a – 6}&{ – 2}\\
{ – 1}&{ – 1}&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
6\\
9\\
b
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
0\\
0\\
0
\end{array}} \right)$

$ \Rightarrow  – 6 – 9 + b = 0 \Rightarrow b = 15$

and $6(10-a)+9(a-6)-2(b)=0$

$ \Rightarrow 60 – 6a + 9a – 54 – 30 = 0$

$ \Rightarrow 3a = 24 \Rightarrow a = 8$

Hence, $a=8,b=15$.