It appears there is no solution to this system of equations.
The fact becomes clear when you try to eliminate variables and wind up with an equation such as 0 = something that is not equal to 0.
Since this equation is not true, the system of equations has no solution.
If you had come up with a row that says something like 0 = 0 which is true, then the system of equations would probably have had a dependent solution which means that the solution depends on the value of one of the variables.
Since you started with a system of 3 equations in 4 unknowns, the best you could have hoped for would have been a dependent solution.
I will go through the solution to show you what happened.
Your equations are:
1x + 1y + 1z + 1u = 1
2x + 3y - 4z + 5u = 2
3x + 4y - 3z + 6u = 0
Subtract 2 times row 1 from row 2 to get:
0x + 1y - 6z + 3u = 0
Subtract 3 times row 1 from row 3 to get:
0x + 1y - 6z + 3u = -3
You are left with 2 equations in 3 unknowns.
They are:
1y - 6z + 3u = 0
1y - 6z + 3u = -3
If you subtract the first equation from the second equation you will get:
0 = -3
Since 0 = -3 is false, this leads to the conclusion that there is no solution to this system of equations.
In order for there to be a unique solution to this system of equations, you would have had to have 4 equations in 4 unknowns.