If $\left|\begin{array}{ccc}{3}^2+k& 4^2& 3^2+3+k\\ 4^2+k& 5^2& 4^2+4+k\\ 5^2+k& 6^2& 5^2+5+k\end{array}\right|=0$, then the value of k is 

Breaking the given determinant into two determinants, we get 

$\begin{array}{l}\left|\begin{array}{ccc}{3}^2+\mathrm{k}& 4^2& 3^2+3+\mathrm{k}\\ 4^2+\mathrm{k}& 5^2& 4^2+4+\mathrm{k}\\ 5^2+\mathrm{k}& 6^2& 5^2+5+\mathrm{k}\end{array}\right|+\left|\begin{array}{ccc}{3}^2+\mathrm{k}& 4^2& 3\\ 4^2+\mathrm{k}& 5^2& 4\\ 5^2+\mathrm{k}& 6^2& 5\end{array}\right|=0\\ \Rightarrow 0+\left|\begin{array}{ccc}9+\mathrm{k}& 16& 3\\ 7& 9& 1\\ 9& 11& 1\end{array}\right|=0\end{array}$

[ Applying ${\mathrm{R}}_3-{\mathrm{R}}_2$ and ${\mathrm{R}}_2-{\mathrm{R}}_1$ in second det.] 

$\left.\Rightarrow \left|\begin{array}{ccc}9+\mathrm{k}& 16& 3\\ 7& 9& 1\\ 2& 2& 0\end{array}\right|=0\text{ [Applying }{\mathrm{R}}_3-{\mathrm{R}}_2\right]$

$\begin{array}{l}\Rightarrow \left|\begin{array}{ccc}9+\mathrm{k}& 7-\mathrm{k}& 3\\ 7& 2& 1\\ 2& 0& 0\end{array}\right|=0\text{ [Applying }{\mathrm{C}}_2-{\mathrm{C}}_1\text{ ] }\\ \Rightarrow 2(7-\mathrm{k}-6)=0\Rightarrow \mathrm{k}=1\end{array}$