Conditional probability problem $P(A,B,C)$ I have a probability definition given as:
$$P(A,B,C)=P(A|B,C)\times P(B) \times P(C)$$
where B and C are independent events. Also, I have values for $P(B)$ , $P(C)$ and $P(A|B,C)$. I can not figure out how to proceed to get the $P(A)$.  

The problem is underdetermined; that is, the values of $P(A|BC),P(B),P(C)$, together with the fact that $B$ and $C$ are independent, are not sufficient to determine $P(A)$. 
Note that $P(ABC) = P(A|BC)P(BC) = P(BC|A)P(A)$, so $p:=P(A)$ and $q:=P(BC|A)$ might be any reals in the interval $[0,1]$ such that 
$pq=P(A|BC)P(B)P(C).$
Example
Here's an explicit example where $B$ and $C$ are independent and $P(B)=P(C)=P(A|BC)=\frac{1}{2}$, yet $P(A)$ may be any value in the interval $[\frac{1}{8},\frac{7}{8}]$: 
Toss a fair coin twice. Let $B=$"Head on 1st toss" and $C=$"Head on 2nd toss". Then $P(BC)=P(B)\,P(C)=\frac{1}{2}\frac{1}{2}=\frac{1}{4}$ and $P(\overline{BC})=1-P(BC)=\frac{3}{4}$. Now make a third toss, using the same fair coin if $BC$ occurred else using a different coin with arbitrary probability $p\in (0,1)$ of yielding a Head. Let $A=$"Head on 3rd toss". Then $P(A)$ depends on the value freely assigned to $P(A\mid \overline{BC})=p$: 
$$\begin{align}P(A)&=P(A\mid BC)\,P(BC)+P(A\mid \overline{BC})\,P(\overline{BC})\\
&=\frac{1}{2}\,\frac{1}{4}+p\,\frac{3}{4}.
\end{align}$$
Also, notice that as $p$ can be freely chosen in the interval $[0,1]$, it happens that $P(BC\mid A)=\frac{1}{8\,P(A)}\in[\frac{1}{7},1]$, whereas $P(BC)=\frac{1}{4}$.
