TITLE: A homogeneous polynomial which is identically equal to 0
QUESTION [5 upvotes]: I totally believe that there should be a proof to the following statement:
Let $A,B$ and $C$ be three homogeneous polynomials in $\mathbb{C}[x,y,z]$ of the same degree $d>1$. And assume that 
$$
xA+yB+zC \equiv 0 \quad (\text{identically equal to } 0)
$$
then all of $A,B$ and $C$ have to have common zero(s) which are not $(x,y,z)=(0,0,0)$.
But I realize that it is not obvious to me. Is there a way to prove the above statement? Or in fact this is not true? Any comment is welcome.

REPLY [6 votes]: Expanding on Mohan's comment.
Suppose the contrary. From Projective Nullstellensatz we have $\sqrt{(A,B,C)}=(x,y,z)$. It follows that the grade of the ideal $(A,B,C)$ equals 3. Since $A,B,C$ are homogeneous they form a regular sequence. In particular, $C$ is a non-zero divisor on $\mathbb C[x,y,z]/(A,B)$ and from $zC\in(A,B)$ we get $z\in(A,B)$, a contradiction.