Breaking the given determinant into two determinants, we get
\(\begin{aligned}
& \left|\begin{array}{lll}
3^2+\mathrm{k} & 4^2 & 3^2+\mathrm{k} \\
4^2+\mathrm{k} & 5^2 & 4^2+\mathrm{k} \\
5^2+\mathrm{k} & 6^2 & 5^2+\mathrm{k}
\end{array}\right|+\left|\begin{array}{lll}
3^2+\mathrm{k} & 4^2 & 3 \\
4^2+\mathrm{k} & 5^2 & 4 \\
5^2+\mathrm{k} & 6^2 & 5
\end{array}\right|=0 \\
& \Rightarrow 0+\left|\begin{array}{ccc}
9+\mathrm{k} & 16 & 3 \\
7 & 9 & 1 \\
9 & 11 & 1
\end{array}\right|=0
\end{aligned}\)
[Applying \(R_3-R_2\) and \(R_2-R_1\) in second det.]
\(\begin{aligned}
& \Rightarrow\left|\begin{array}{ccc}
9+\mathrm{k} & 16 & 3 \\
7 & 9 & 1 \\
2 & 2 & 0
\end{array}\right|=0\left[\text { Applying } \mathrm{R}_3-\mathrm{R}_2\right] \\
& \left.\Rightarrow\left|\begin{array}{ccc}
9+\mathrm{k} & 7-\mathrm{k} & 3 \\
7 & 2 & 1 \\
2 & 0 & 0
\end{array}\right|=0 \text { [Applying } \mathrm{C}_2-\mathrm{C}_1\right] \\
& \Rightarrow 2(7-\mathrm{k}-6)=0 \Rightarrow \mathrm{k}=1
\end{aligned}\)