Solve:-
$x-2y+3z=11$
$3x+y-z=2$
$5x+3y+2z=3$

Given,

$x-2y+3z=11-----\left(1\right)$

$3x+y-z=2------\left(2\right)$

$5x+3y+2z=3-----\left(3\right)$

Multiplying (1) with 3 and then solving with (2),we get,

$10z-7y=31-----\left(4\right)$

Multiplying (1) with 5 and then solving with (3),we get,

$z-y=4------\left(5\right)$

Now,multiplying (5) with 7 and then solving with (4),we get,

$=>z=1$

Putting the value of $z$ in (5),

$y=-3$

putting the value of $z$ and $y$ in (1),

$x=2$