The system of linear equations \( x+y+z=6, x+2 y+3 z=10 \) and \( x+2 y+a z=b \) has no
solutions when

Given equations,
\(x+y+z=6 \rightarrow(1)\)
\(x+2 y+3 z=10 \rightarrow(2)\)
\(x+2 y+a z=b \rightarrow(3)\)
\(\left[\begin{array}{llll}1 & 1 & 1 \\ 1 & 2 & 3\end{array}\right]\left[\begin{array}{l}x \\ y \\ 1 & 2 & a\end{array}\right]=\left[\begin{array}{c}6 \\ 10\end{array}\right]\)
Now \(\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3\end{array}\right]=\left[\begin{array}{c}6 \\ 10\end{array}\right]\)
\(R_{3} \rightarrow R_{3}-R_{2}\), We have
\(\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 0 & 0 & a-3\end{array}\right]=\left[\begin{array}{c}10 \\ b\end{array}\right]=10\)
For now solutions \(a-3=0 \Rightarrow=a=3\) and \(b-10 \neq 0 \Rightarrow b \neq 10\)
Alternating Solution
Comparing Eqs. ( 2 ) and (3), we have
\(a=3, b=10\)
So, if \(a=3, b \neq 10\) then the system has no solution.