Since the right side of each equation is 0, it is trivial that $x_1= x_2= x_3= x_4= x_5= 0$ is a solution. But since there are three equations in five unknowns, there exist other, non trivial solutions.
One way to find all possible solutions is to start trying to solve the equations. For example, if we multiply the first equation by 2 and subtract from the second equation we eliminate both $x_1$ and $x_5$: $-8x_2+ 7x_3- 25x_4= 0$. If we multiply the third equation by 2 and subtract from the second equation, we eliminate $x_1$ again: $14x_2- 7x_3+ 25x_4- 4x_5= 0$. And if we add those two equations, we eliminate both $x_3$ and $x_4$: 6x_2- 4x_5= 0. From that $x_5= \frac{3}{2}x_2$. Putting that into $14x_2- 7x_3+ 25x_4- 4x_5= 0$ we get $8x_2- 7x_3+ 25x_1= 0$. Solving that for $x_1$, $x_1= \frac{7}{25}x_3- \frac{8}{25}x_2$. We are through with the equations so this is as far as we can go. writing "A" for $x_3$ and "B" for $x_2$, $x_1= \frac{7}{25}A- \frac{8}{25}B$, $x_2= A$, $x_4= B$, and you can solve for $x_3$ and $x_5$ in terms of A and B.