Solve the following quadratic equations by factorization:

We have been given

$\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x}$

$x^{2}-x+2 x^{2}-4 x=6\left(x^{2}-x-2 x+2\right)$

$3 x^{2}-13 x+12=0$

Therefore,

$3 x^{2}-9 x-4 x+12=0$

$3 x(x-3)-4(x-3)=0$

$(3 x-4)(x-3)=0$

Therefore,

$3 x-4=0$

$3 x=4$

$x=\frac{4}{3}$

or,

$x-3=0$

$x=3$

Hence, $x=\frac{4}{3}$ or $x=3$.