Answer
$(1,1,\frac{1}{3})$
Work Step by Step
$2x-y+3z = 2$ Equation $(1)$
$x+y -6z=0$ Equation $(2)$
$3x+4y-3z=6$ Equation $(3)$
Adding Equation $(1)$ and Equation $(2)$
$2x+x-y+y+3z-6z=2+0$
$3x -3z = 2$ Equation $(4)$
Multiplying Equation $(1)$ by $4$ and adding with Equation $(3)$
$8x+3x-4y+4y+12z-3z = 8 +6$
$11x +9z = 14$ Equation $(5)$
Multiplying Equation $(4)$ by $3$ and adding with Equation $(5)$
$9x+11x-9z+9z = 6+14$
$20x = 20$
$x= 1$
Substituting $x$ value in Equation $(4)$
$3(1)-3z=2$
$3-3z = 2$
$-3z=2-3$
$-3z=-1$
$3z=1$
$z=\frac{1}{3}$
Substituting $x$ and $z$ values in Equation $(2)$
$x+y -6z=0$
$1+y -6(\frac{1}{3})=0$
$1+y-2=0$
$y-1=0$
$y=1$
Solution: $(1,1,\frac{1}{3})$