Question: > 
> Find the solutions of the equation $$\frac{a^2(x-b)(x-c)}{(a-b)(a-c)}+\frac{b^2(x-a)(x-c)}{(b-a)(b-c)}+\frac{c^2(x-a)(x-b)}{(c-a)(c-b)}=3x-2$$
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If we consider the expression as an identity and substitute $x=a$ we get the value of $a=1,2$ similarly $b=1,2$ and $c=1,2$  

If we substitute these values into the expression at least one of the denominator will be equal to $0$ which is not possible therefore this expression is not an identity. How to proceed further?

Answer: 
Assume $a,b,c$ are fixed distinct (all different) real number values. (Otherwise the left-hand side is undefined.)

Then the left-hand side is a quadratic function in $x$. Call this function $f(x)$.

We have $f(a) = a^2$, $f(b)=b^2$ and $f(c)=c^2$. Since a quadratic is determined by its values at three points, we must have $f(x)=x^2$. (This expression for $f(x)$ should be independent of the values of $a, b$, and $c$, provided, as noted above, that these values are all different.)

So the equation becomes $x^2=3x-2$, which you can now easily solve for $x$.
Add $-x^2$ on both sides , Let expression on LHS be $f(x)$

Therefore LHS will be a quadratic in $x$ will be become zero three times clearly as

$f(a)=f(b)=f(c)=0$ Now if a quadratic equation has more than three roots, then all complex numbers will be it's roots and therefore it will identically be equal to zero.

$f(x)\equiv 0$ and therefore since $f(x)=3x-2-x^2$

$3x-2-x^2\equiv 0$

which is possible for $x=1,2$ only