Form the matrix and compute the determinant: \(\begin{vmatrix}1&2&3\\2&5&7\\1&3&5\end{vmatrix} = 1(25-21)-2(10-7)+3(6-5) = 4-6+3=1 \neq 0\). Wait: \(= 1(25-21)-2(10-7)+3(6-5)=4-6+3=1\). Det\(\neq 0\) means linearly independent. Actually recomputing: \(5\times5-7\times3=25-21=4\); \(2\times5-7\times1=10-7=3\); \(2\times3-5\times1=6-5=1\). Det\(=1(4)-2(3)+3(1)=4-6+3=1\neq0\). Linearly independent.