Step 1: Solve by elimination as an independent cross-check instead of the matrix formula:
From eq1: $x=7+y-2z$. Substitute into eq2: $3(7+y-2z)+4y-5z=-5\Rightarrow21+3y-6z+4y-5z=-5\Rightarrow7y-11z=-26$.
Step 2: Substitute the same expression for x into eq3:
$2(7+y-2z)-y+3z=12\Rightarrow14+2y-4z-y+3z=12\Rightarrow y-z=-2\Rightarrow y=z-2$.
Step 3: Substitute $y=z-2$ into the reduced equation $7y-11z=-26$:
$7(z-2)-11z=-26\Rightarrow7z-14-11z=-26\Rightarrow-4z=-12\Rightarrow z=3$. Then $y=3-2=1$, and $x=7+1-6=2$.
Step 4: Verify in the original third equation:
$2(2)-1+3(3)=4-1+9=12$ ✓, matching the given RHS.
Final Answer:
Elimination confirms the same solution as the matrix method.
\[ \boxed{x=2,\ y=1,\ z=3} \]