If $$3yz + 2y + z + 1 = 0$$ and $$3zx + 2z + x + 1 = 0$$, then prove that $$3xy + 2x + y + 1 = 0$$.

PLACEHOLDER ANSWER - this is a proof question. From the first equation $$z = -\frac{2y + 1}{3y + 1}$$, so $$3z + 1 = -\frac{3y + 2}{3y + 1}$$ and $$2z + 1 = -\frac{y + 1}{3y + 1}$$. From the second, $$x = -\frac{2z + 1}{3z + 1} = -\frac{y + 1}{3y + 2}$$. Then $$3xy + 2x + y + 1 = x(3y + 2) + y + 1 = -(y + 1) + y + 1 = 0$$.