Attempt to eliminate $z$ from two pairs of equations. M1
Combining (1) and (2): Multiply (2) by 2 and add to (1):
$(x - y + 2z) + 2(2x + 3y - z) = 0 + 2(9)$
$x - y + 2z + 4x + 6y - 2z = 18$
$5x + 5y = 18 \Rightarrow x + y = \frac{18}{5}$ (Eq. 4) A1
Combining (1) and (3): Multiply (1) by 2 and subtract (3):
$2(x - y + 2z) - (-3x + y + 4z) = 0 - 5$
$(2x - 2y + 4z) + 3x - y - 4z = -5$
$5x - 3y = -5$ (Eq. 5) A1
Attempt to solve the system of two variables (Eq. 4 and Eq. 5). M1
From (Eq. 4), $x = \frac{18}{5} - y$. Substitute into (Eq. 5):
$5(\frac{18}{5} - y) - 3y = -5$
$18 - 5y - 3y = -5$
$18 - 8y = -5$
$8y = 23 \Rightarrow y = \frac{23}{8}$ A1
Substitute $y = \frac{23}{8}$ into $x = \frac{18}{5} - y$ to find $x$:
$x = \frac{18}{5} - \frac{23}{8} = \frac{144}{40} - \frac{115}{40} = \frac{29}{40}$ A1
Substitute $x$ and $y$ into (1) to find $z$:
$2z = y - x = \frac{115}{40} - \frac{29}{40} = \frac{86}{40}$
$z = \frac{43}{40}$ A1
State the intersection point: $(\frac{29}{40}, \frac{23}{8}, \frac{43}{40})$ A1

8 marks total