May,
You can solve it in 3 ways by repeated: variable substitution to reduce # of variables, equation multiplication/addition to eliminate a variable, and the matrix approach which is related to the second method. I will use the third method, but first I will move equation 1 down to be equation 3 so I don't have to deal with fractions. I want to begin with the coefficient of x to be 1. From the 3 equations form the coefficient & constants matrix as follows:
| x-coeff y-coeff z-coeff Constant after = sign |
| +1 -3 +4 -14 |
| +1 -2 -3 -11 |
| +2 +1 +1 - 7 |
Now multiply row 1 coefficients appropriately (with sign & magnitude) and add multiplied row 1 to make the remaining non-row 1 coefficients in column 1 both zero. You get the modified matrix:
| +1 -3 +4 -14 |
| 0 +1 -7 +3 |
| 0 +7 -7 +21|
Now do the same thing by using row 2 make the other column 2 entries in rows 1 & 3 both zero, and get:
| +1 0 -17 -5 |
| 0 +1 -7 +3 |
| 0 0 +42 0|
Finally, divide row 3 by 42 to make the pivot entry 1, then non-row 3 column 3 entries zero, and get:
| +1 0 0 -5 |
| 0 +1 0 +3 |
| 0 0 +1 0 |
You've made the x,y,z orthogonal and the answer to your system of equations is in the fourth column:
x = -5; y = +3; and z = 0. Confirm this answer by plugging these values into the 3 original equations.
NOTE. You can use this method for solving a system of n equations with n independent unknowns. If some of the equations are multiples or linear combinations of other equations, then they're not all independent. You have fewer independent equations than unknown variables, and insufficient info to get a unique solution for all variables.
