Explanation

(b)
Applying $C_{1} \rightarrow C_{1} + C_{2},\$we get
$\left| \begin{matrix}
\ ^{10}C_{4} + \ ^{10}C_{5} & \ ^{10}C_{5} & \ ^{11}C_{m} \\
\ ^{11}C_{6} + \ ^{11}C_{7} & \ ^{11}C_{7} & \ ^{12}C_{m + 2} \\
\ ^{12}C_{8} + \ ^{12}C_{9} & \ ^{12}C_{9} & \ ^{13}C_{m + 4}
\end{matrix} \right| = 0$
$\Rightarrow \left| \begin{matrix}
\ ^{11}C_{5} & \ ^{10}C_{5} & \ ^{11}C_{m} \\
\ ^{12}C_{7} & \ ^{11}C_{7} & \ ^{12}C_{m + 2} \\
\ ^{13}C_{9} & \ ^{12}C_{9} & \ ^{13}C_{m + 4}
\end{matrix} \right| = 0$
It means either two rows or two columns are identical.
$\therefore\ ^{11}C_{5} = \ ^{11}C_{m},\ \ ^{12}C_{7} = \ ^{12}C_{m + 2},\ \ ^{13}C_{9} = \ ^{13}C_{m + 4}$
$\Rightarrow \ \ m = 5$