Solution of $x-y+z=4 ; x-2 y+2 z=9$ and $2 x+y+3 z=1$ is

The given system of linear equations is:
(1) $x - y + z = 4$
(2) $x - 2y + 2z = 9$
(3) $2x + y + 3z = 1$

Subtract equation (1) from equation (2):
$(x - 2y + 2z) - (x - y + z) = 9 - 4$
$-y + z = 5$ --- (4)

Multiply equation (1) by 2 and subtract from equation (3):
$(2x + y + 3z) - 2(x - y + z) = 1 - 2(4)$
$2x + y + 3z - 2x + 2y - 2z = 1 - 8$
$3y + z = -7$ --- (5)

Subtract equation (4) from equation (5):
$(3y + z) - (-y + z) = -7 - 5$
$4y = -12$
$y = -3$

Substitute $y = -3$ into equation (4):
$-(-3) + z = 5$
$3 + z = 5$
$z = 2$

Substitute $y = -3$ and $z = 2$ into equation (1):
$x - (-3) + 2 = 4$
$x + 3 + 2 = 4$
$x + 5 = 4$
$x = -1$

The solution is $x = -1, y = -3, z = 2$.

Answer: $x=-1 ; \quad y=-3 ; \quad z=2$