The quadratic equations $x^2+\left(a^2-2\right)x-2a^2=0$ and $x^2-3x+2=0$ have

Let $a$ be a common root of the equations

$x^2+\left(a^2-2\right)x-2a^2=0$ and $x^2-3x+2=0$

Then,

${\alpha }^2+\left(a^2-2\right)\alpha -2a^2=0$ and ${\alpha }^2-3\alpha +2=0$

Now,

${\alpha }^2-3\alpha +2=0\Rightarrow \alpha =1 ,2$

Putting $\alpha =1$ in ${\alpha }^2+\left(a^2-2\right)\alpha -2a^2=0$, we get 

$\Rightarrow a^2+1=0$, which is not possible for any $a\in R$.

Putting $\alpha =2$ in ${\alpha }^2+\left(a^2-2\right)\alpha -2a^2=0$, we get

$4+2\left(a^2-2\right)-2a^2=0,$ which is true for all $a\in R$.

Thus, the two equations have exactly one common root for all  $a\in R$.