$\left|\begin{array}{ccc}2 & 3 & 5 \\ 3 & 5 & 2 \\ 5 & 2 & 3\end{array}\right|+\left|\begin{array}{ccc}1 & 1 & 1 \\ 7 & 11 & 13 \\ 49 & 121 & 169\end{array}\right|$
We know that
$$
\begin{aligned}
& \left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|=3 a b c-a^3-b^3-c^3 \\
& \Rightarrow\left|\begin{array}{lll}
1 & 1 & 1 \\
a & b & c \\
a^2 & b^2 & c^2
\end{array}\right|=(a-b)(b-c)(c-a) \\
& \therefore\left|\begin{array}{lll}
2 & 3 & 5 \\
3 & 5 & 2 \\
5 & 2 & 3
\end{array}\right|=3(2)(3)(5)-2^3-3^3-5^3 \\
& =90-8-27-125=-70 \\
& \left|\begin{array}{ccc}
1 & 1 & 1 \\
7 & 11 & 13 \\
49 & 121 & 169
\end{array}\right|=(7-11)(11-13)(13-7) \\
& =-4 \times(-2) \times 6=48 \\
& \therefore \text { Required sum }=-70+48=-22 .
\end{aligned}
$$