The given system of equations is:

1) x + y + z = 0
2) 2x - y - 3z = 0
3) 3x - 5y + 4z = 0

To determine the number of solutions, we can use the method of elimination or substitution.

Using the Method of Elimination:

To eliminate variables, we can start by multiplying the first equation by 2, and the second equation by 3. This will allow us to eliminate the x variable when we subtract the equations.

Multiplying the first equation by 2:
2(x + y + z) = 2(0)
2x + 2y + 2z = 0

Multiplying the second equation by 3:
3(2x - y - 3z) = 3(0)
6x - 3y - 9z = 0

Now we can subtract the multiplied equations to eliminate the x variable:
(6x - 3y - 9z) - (2x + 2y + 2z) = 0 - 0
6x - 2x - 3y - 2y - 9z - 2z = 0
4x - 5y - 11z = 0 --> (Equation 4)

Now we have a new equation (Equation 4) involving only the variables y and z. We can use this equation along with the original third equation to eliminate one more variable.

Multiplying the third equation by 4:
4(3x - 5y + 4z) = 4(0)
12x - 20y + 16z = 0

Now we can add the multiplied equation to the third equation to eliminate the z variable:
(12x - 20y + 16z) + (3x - 5y + 4z) = 0 + 0
12x + 3x - 20y - 5y + 16z + 4z = 0
15x - 25y + 20z = 0 --> (Equation 5)

Now we have a new equation (Equation 5) involving only the variable x and y. We can use this equation along with the original first equation to eliminate one more variable.

Multiplying the first equation by 15:
15(x + y + z) = 15(0)
15x + 15y + 15z = 0

Now we can subtract the multiplied equation from the fifth equation to eliminate the y variable:
(15x - 25y + 20z) - (15x + 15y + 15z) = 0 - 0
15x - 15x - 25y - 15y + 20z - 15z = 0
-40y + 5z = 0 --> (Equation 6)

Now we have an equation (Equation 6) involving only the variable y and z. Let's examine the system of equations that we have obtained so far:

1) x + y + z = 0
4) 4x - 5y - 11z = 0
5) 15x - 25y + 20z