(3) $$ \begin{aligned} x+y+z & =6 \ 2 x-y+2 z & =3 \ -x+3 y-z & =-4\end{aligned} $$

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Solve the system of equations $$ x+y+z=6;2x-y+2z=3;-x+3y-z=-4 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}x+y+z=6\2x-y+2z=3\-x+3y-z=-4\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=6-y-z\2x-y+2z=3\-x+3y-z=-4\end{array}ight.$$
- step2: Substitute the value of $$x:$$
  $$\left\{ \begin{array}{l}2\left(6-y-zight)-y+2z=3\-\left(6-y-zight)+3y-z=-4\end{array}ight.$$
- step3: Simplify:
  $$\left\{ \begin{array}{l}12-3y=3\-6+4y=-4\end{array}ight.$$
- step4: Solve the equation:
  $$\left\{ \begin{array}{l}y=3\-6+4y=-4\end{array}ight.$$
- step5: Substitute the value of $$y:$$
  $$-6+4	imes 3=-4$$
- step6: Simplify:
  $$6=-4$$
- step7: Calculate:
  $$\left\{ \begin{array}{l}x \in \varnothing \y \in \varnothing \z \in \varnothing \end{array}ight.$$
- step8: Rewrite:
  $$(x, y, z) \in \varnothing$$
El sistema de ecuaciones dado no tiene solución.
El sistema de ecuaciones no tiene solución.

(3)

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