Question

If $D = \left| \begin{matrix} 1 & 1 & 1 \\ 2 & 5 & a \\ 1 & 2 & 3 \end{matrix} \right| = 0 \Rightarrow a = 8$

If $D_{1} = \left| \begin{matrix} 6 & 1 & 1 \\ 36 & 5 & a \\ b & 2 & 3 \end{matrix} \right| = 0 \Rightarrow ab - 5\text{ }b - 12a + 54 = 0$

If $D_{2} = \left| \begin{matrix} 1 & 6 & 1 \\ 2 & 36 & a \\ 1 & b & 3 \end{matrix} \right| = 0 \Rightarrow ab - 6a - 2b - 36 = 0$

If $D_{3} = \left| \begin{matrix} 1 & 1 & 6 \\ 2 & 5 & 36 \\ 1 & 2 & \text{ }b \end{matrix} \right| = 0 \Rightarrow \text{ }b = 14$

For $a = 8\&\text{ }b = 14 \Rightarrow D_{1}\& D_{2}$ are also zero

For $a = 8\&\text{ }b = 14 \Rightarrow D = D_{1} = D_{2} = D_{3} = 0$

⇒ infinitely many solutions.