Sequences; the binomial theorem

Question

Find $A, B, C$ such that the expansion in ascending powers of $x$ of $A(1+x)^7 +B(1-2x)^5 +C(1-3x)^3$ begins with terms in $x^2$

I put $A+B+C=0$ and $ 7A -10B -9C=0$ to let $x^2$ be the first term in the expansion. So $A, B, C$ could be $-k, 16k, -17k$ or $k, 16k, -17k$
or $-k/272, -k/17, k/16$ or $k/272, k/17, -k/16$

Does it mean $A, B ,C$ can be any of these.

Answer 1

HINT

Let apply for each term $$(x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k $$

to find the $x^2$ coefficient, then set the condition for A,B,C such that the condition is satisfied.