intersection

6x^2-5x+1<0 \Rightarrow \left(2x-1 \right)\left(3x-1 \right)<0
\Rightarrow x\in \left(\frac{1}{3}, \frac{1}{2}\right)

3x^2+2x-1>0 \Rightarrow \left(x+1 \right)\left(3x-1 \right)
\Rightarrow x\in \left(-\infty,-1 \right)\cup \left(\frac{1}{3},\infty \right)

2x^2+x-3<0 \Rightarrow \left(2x+3 \right)\left(x-1 \right)<0
\Rightarrow x\in \left(-\frac{3}{2},1 \right)

Now, the required intersection is:

\left(\frac{1}{3},\frac{1}{2}\right)\cap\left(\left(-\infty,-1 \right)\cup \left(\frac{1}{3},\infty \right) \right)\cap\left(-\frac{3}{2},1 \right)=\left(\frac{1}{3},\frac{1}{2} \right)

But the answer is different....!!