Question 2108
22x+5y+7z=12 --- (1)
10x+3y+2z=5  --- (2)
9x+2y+12z=14 --- (3)

Eqn(1) x 3
66x+15y+21z=36 --- (4)

eqn(2) x 5
50x+15y+10z=25 --- (5)

Eqn(4) - eqn(5)
16x+11z=11  --- (6)

Eqn(2) x 2
20x+6y+4z=10  --- (7)

eqn(3) x 3
27x+6y+36z=42  --- (8)

Eqn(8) - eqn(7)
7x+32z=32  --- (9)

Eqn(6) x 7 gives
112x+77Z=77 --- (10)

Eqn(9) x 16
112x+512z=512 --- (11)

eqn(11) - eqn(10)
435z=435 => z = 1

Substituting the value of z = 1 in eqn(9) gives
7x+32=32 => x = 0

10x+3y+2z=5  --- (2)

Substituting the value of x = 0 and z = 1 in eqn(2) gives
10.0+3y+2.1=5 => 3y = 3 => y = 1


x=0, y=1, z=1