Let $g\left( x \right) = \frac{{a{x^3}}}{3} + b.\frac{{{x^2}}}{2} + cx$

      $g'\left( x \right) = a{x^2} + bx + c$

given:$a{x^2} + bx + c = 0$ and $2a + 3b + 6c = 0$

Statement – $2$:

$(i)   g(0) =0$ and $g(1)$

                                       $ = \frac{a}{3} + \frac{b}{2} + c = \frac{{2a + 3b + 6c}}{6}$

                                       $ = \frac{0}{6} = 0$

$ \Rightarrow g\left( 0 \right) = g\left( 1 \right)$

$(ii)$ $g$ is continuose on $[0,1]$ and differentiable on $(0,1)$

$\therefore $ By Rolle's theorem $\exists k \in \left( {0,1} \right)$such that $g'(k)=0$

This holds the statement $2$. Also, from statement – $2$, we can say $a{x^2} + bx + c = 0$ has at least one root in $(0,1)$

Thus, statement – $1$ and $2$ both are true and statement – $2$ is a correct explantion for statement – $1$.