Given, $a, b$ and $c$ are the roots of equation
$\begin{aligned}
& x^3+q x+r=0 \text {, } \\
& \therefore \quad a+b+c=0 \\
& a b+b c+c a=q \\
& \text { and } \\
& a b c=-r \\
& \because \quad a+b+c=0 \\
& \Rightarrow \quad(a+b+c)^2=0 \\
& \Rightarrow a^2+b^2+c^2+2 a b+2 b c+2 c a=0 \\
& \Rightarrow \quad a^2+b^2+c^2+2(a b+b c+c a)=0 \\
& \Rightarrow \quad a^2+b^2+c^2=-2 q \\
&Now, (a-b)^2+(b-c)^2+(c-a)^2\\
& =a^2+b^2-2 a b+b^2+c^2-2 b c+c^2+a^2-2 ac\\
& =2 a^2+2 b^2+2 c^2-2(a b+b c+c a) \\
& =2\left(a^2+b^2+c^2\right)-2(a b+b c+c a) \\
& =2(-2 q)-2(q) &[by Eq. (i)] \\
& =-4 q-2 q \\
& =-6 q \text {. } \\
&
\end{aligned}$