In this above question,

we know that,

$$\begin{gathered} = a^3+b^3+c^3-3abc=(a+b+c)\times \frac{1}{2}[ \\ {\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(a-c\right)}^2] ---(1) \end{gathered}$$

Put the value of (a+b+c)=0 from equation (1)

$$\begin{gathered} = a^3+b^3+c^3-3abc=0\times \frac{1}{2}[ \\ {\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(a-c\right)}^2] \\ =a^3+b^3+c^3-3abc=0 \\ =a^3+b^3+c^3=3abc \end{gathered}$$

∴ Given expression in the question is

$$\begin{gathered} = \frac{{\left(m-n\right)}^3+{\left(n-r\right)}^3+{\left(r-m\right)}^3}{6\left(m-n\right)\left(n-r\right)\left(r-m\right)} \\ = \frac{\left(a^3+b^3+c^3\right)}{6abc} \\ =\frac{3abc}{6abc}=\frac{1}{2} \end{gathered}$$

Hence, Option A is correct.