The second vector that's perpendicular to the line we want (call it L) is missing; I'll denote it by ⟨a, b, c⟩.
The cross product of ⟨1, 1, 0⟩ and ⟨a, b, c⟩ is perpendicular to both of these vectors. The direction vector of L will be parallel to this cross product.
Compute the cross product.
⟨1, 1, 0⟩ × ⟨a, b, c⟩ = ⟨1•c - 0•b, 0•a - 1•c, 1•b - 1•a⟩ = ⟨c, -c, b - a⟩
Then the vector equation of L is
L(t) = ⟨c, -c, b - a⟩ t + ⟨-3, 3, -1⟩
As parametric equations, we write
x(t) = ct - 3
y(t) = -ct + 3
z(t) = (b - a) t - 1
Eliminating the parameter above, we get
x + y = (ct - 3) + (-ct + 3) = 0 ⇒ x = -y
and substituting
x = ct - 3 ⇒ t = (x + 3)/c
into z(t), we get
z = (b - a) (x + 3)/c - 1 ⇒ x = c (z + 1)/(b - a) - 3
As symmetric equations, then, we write
x = -y = c (z + 1)/(b - a) - 3