#### Answer

$(x,y,z)=(4,3,4)$

#### Work Step by Step

Given: $x-2y+3z=10$ ; $2y-z=2; 3z=12$
Here $3z=12 \implies z=4$
Plug $z=4$ in the second equation.
we get $2y-4=2 \implies y=3$
Now, Plug $y=3,z=4$ in the first equation.
we have $x-2(3)+3(4)=10$
or, $x=10-6 \implies x=4$
Hence, the solution is:$(x,y,z)=(4,3,4)$