Solve the following system of linear equations:
$x+y+z=15$, $2x-y+z=12$ and $x-y-z=-1$

The correct option is x = 7, y = 5, z = 3.

We are given the following system of three linear equations:

(1) $x+y+z=15$

(2) $2x-y+z=12$

(3) $x-y-z=-1$

Step 1: Add Equation (1) and Equation (3) to eliminate variables y and z.

($x+y+z$) + ($x-y-z$) = $15+(-1)$

$2x=14$

$x=7$

Step 2: Add Equation (2) and Equation (3) to eliminate variable y.

($2x-y+z$) + ($x-y-z$) = $12+(-1)$

$3x-2y=11$

Substitute $x=7$ into this equation:

$3\left(7\right)-2y=11$

$21-2y=11$

$2y=21-11$

$2y=10$

$y=5$

Step 3: Substitute the values of x and y into Equation (1) to find z.

$7+5+z=15$

$12+z=15$

$z=15-12$

$z=3$

Thus, the solution to the system of equations is x = 7, y = 5, z = 3.