We are asked to show that:
$$ \det\left[\begin{array}{rrr} 2 & 3 & 7 & 1 & 3\\ 2 & 3 & 7 & 1 & 5\\ 2 & 3 & 6 & 1 & 9\\ 4 & 6 & 2 & 3 & 4\\ 5 & 8 & 7 & 4 & 5 \end{array}\right] = 2 $$
By using suitable elementary row and column operations as well as row and column expansions. I have used:
$$ R_2 - R_1 \to R_2 $$ $$ R_3 - R_1 \to R_3 $$ $$ 2R_1 - R_4 \to R_4 $$
Which is the following matrix:
$$ \det\left[\begin{array}{rrr} 2 & 3 & 7 & 1 & 3\\ 0 & 0 & 0 & 0 & 2\\ 0 & 0 & -1 & 0 & 6\\ 0 & 0 & 12 & -1 & 2\\ 5 & 8 & 7 & 4 & 5 \end{array}\right] = 2 $$
I would surmise now using either a row or column expansion, but need clarification/confirmation.
Expand by row 2, then by (original) row 3, then by row 4, leaving a single $2\times 2$ determinant that you can do manually.