If the system of linear equations $ \begin{aligned} &x+a y+z=3 \\ &x+2 y+2 z=6 \\ &x+5 y+3 z=b \end{aligned}…
If the system of linear equations
$
\begin{aligned}
&x+a y+z=3 \\
&x+2 y+2 z=6 \\
&x+5 y+3 z=b
\end{aligned}
$
has no solution, then
$a=1, b \neq 9$
$a \neq-1, b=9$
$a=-1, b=9$
$a=-1, b \neq 9$
Solution
As the system of equations has no solution then $\Delta$ should be zero and at least one of $\Delta_1$, $\Delta_2$ and $\Delta_3$ should not be zero.
$
\begin{aligned}
&\therefore \Delta=\left|\begin{array}{lll}
1 & a & 1 \\
1 & 2 & 2 \\
1 & 5 & 3
\end{array}\right|=0 \\
&\Rightarrow-a-1=0 \Rightarrow a=-1 \\
&\Delta_2=\left|\begin{array}{lll}
1 & 3 & 1 \\
1 & 6 & 2 \\
1 & b & 3
\end{array}\right| \neq 0 \\
&\Rightarrow b \neq 9
\end{aligned}
$