x + y + z =3 ........(let this be equation 1)

x +2y +3z =4........(let this be equation 2)

x + 4y + 9z =6........(let this be equation 3)

(y,z) = (1,0)

From equation 1 (we get

x = 3-y-z

Substitute value of x  equation  2 & 3 we get:

y + 2z = 1.........(let this be equation 4)

3y =3

respectively.

Hence

y=3/3

substituting  value of y in equation 4 we get z =0

HENCE

y=1

z=0

(y,z) = (1,0)