$$\left\{\begin{array}{lcr}(A)\;\phantom{1}x-3y+6z&=&21\\
(B)\;3x+2y-5z&=&-30\\
(C)\;2x-5y+2z&=&-6\end{array}\right.$$
is equivalent (through $(B)\mapsto(B-3A)$ and $(C)\mapsto (C-2A)$) to$$\left\{\begin{array}{lcr}(A)\;\phantom{1}x-3y+6z&=&21\\
(B)\;\phantom{3x+}11y-23z&=&-93\\
(C)\;\phantom{2x-\,\,\,}y-10z&=&-48\end{array}\right.$$
that is equivalent (through $(B)\mapsto(B-11C)$) to
$$\left\{\begin{array}{lcr}(A)\;\phantom{1}x-3y+6z&=&21\\
(B)\;\phantom{3x+11y-}87z&=&435\\
(C)\;\phantom{2x-\,\,\,}y-10z&=&-48\end{array}\right.$$
Now use $(B)$ to find the value of $z$, plug it into $(C)$ to find the value of $y$, plug them both into $(A)$ to find the value of $x$. This is known as the Gaussian elimination method. A good alternative for $3\times 3$ systems is Cramer's rule. They both lead to:
$$ (x,y,z)=\color{red}{(-3,2,5)} $$