What is the solution to the following system of linear equations?
$$3x-9y+4z=5$$
$$2x+7y+z=12$$
$$3x-z=0$$

Given equations:$3x-9y+4z=5$$2x+7y+z=12$$3x-z=0⟹z=3x$ Substitute $z=3x$ into equations (1) and (2):From equation (1):
$$3x-9y+4(3x)=5$$
$$3x-9y+12x=5$$
$$15x-9y=5\text{— (4)}$$
From equation (2):
$$2x+7y+3x=12$$
$$5x+7y=12\text{— (5)}$$
Now solve equations (4) and (5). Multiply equation (5) by $3$:
$$15x+21y=36\text{— (6)}$$
Subtract equation (4) from equation (6):
$$(15x+21y)-(15x-9y)=36-5$$
$$30y=31$$
$$y=\frac{31}{30}$$
Substitute $y=\frac{31}{30}$ into equation (5):
$$5x+7\left(\frac{31}{30}\right)=12$$
$$5x+\frac{217}{30}=12$$
$$5x=12-\frac{217}{30}$$
$$5x=\frac{360-217}{30}$$
$$5x=\frac{143}{30}$$
$$x=\frac{143}{150}$$
Now substitute the value of $x$ to find $z$:
$$z=3x=3\left(\frac{143}{150}\right)=\frac{143}{50}$$
Thus, the unique solution is:
$$x=\frac{143}{150},y=\frac{31}{30},z=\frac{143}{50}$$