The remainders of the polynomial  $f\left(x\right)$ when divided by  $x+1,x+2,x-2$  are  $6,15,3$  the remainder of $f\left(x\right)$   when divided by $(x+1)(x+2)(x-2)$  is

 $$\begin{gathered} f(-1)=6,f(-2)=15,f\left(2\right)=3 \\ f\left(x\right)=(x+1)(x+2)(x-3)q\left(x\right)+(a{x}^2+bx+c) \end{gathered}$$
Put  $x=-1\Rightarrow a-b+c=6\to \left(1\right)$
$$\begin{gathered} x=-2\Rightarrow 4a-2b+c=15\to \left(2\right) \\ x=2\Rightarrow 4a-2b+c=3\to \left(3\right) \end{gathered}$$        
Solving (1), (2) and (3) we get 
a = 2, b = -3,  c = 1 
Remainder is  $2x^2-3x+1$