If ${\mathrm{\Delta }}_{\mathrm{r}}=\left|\begin{array}{ccc}\mathrm{r}& 612& 915\\ 101{\mathrm{r}}^2& 2\mathrm{r}& 3\mathrm{r}\\ \mathrm{r}& \frac{1}{\mathrm{r}}& \frac{1}{{\mathrm{r}}^2}\end{array}\right|$, then the value of $\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \frac{1}{{\mathrm{n}}^3}\sum _{\mathrm{r}=1}^{\mathrm{n}} {\mathrm{\Delta }}_{\mathrm{r}}$ is _________.

$\begin{array}{c}{\mathrm{\Delta }}_{\mathrm{r}}=\left|\begin{array}{ccc}\mathrm{r}& 612& 915\\ 101{\mathrm{r}}^2& 2\mathrm{r}& 3\mathrm{r}\\ \mathrm{r}& \frac{1}{\mathrm{r}}& \frac{1}{{\mathrm{r}}^2}\end{array}\right|\\ ={\mathrm{r}}^2\left|\begin{array}{ccc}1& 612& 915\\ 101& 2& 3\\ 1& \frac{1}{\mathrm{r}}& \frac{1}{{\mathrm{r}}^2}\end{array}\right| \left(\text{Taking }\mathrm{r}\text{ common from }{\mathrm{C}}_1\text{ and }{\mathrm{R}}_2\right)\\ =\left|\begin{array}{ccc}1& 612& 915\\ 101& 2& 3\\ {\mathrm{r}}^2& \mathrm{r}& 1\end{array}\right|\end{array}$

$\therefore \underset{\mathrm{n}\to \mathrm{\infty }}{lim} \frac{1}{{\mathrm{n}}^3}\sum _{\mathrm{r}=1}^{\mathrm{n}} \left|\begin{array}{ccc}1& 612& 915\\ 101& 2& 3\\ {\mathrm{r}}^2& \mathrm{r}& 1\end{array}\right|$

                 $\begin{array}{c}=\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \frac{1}{{\mathrm{n}}^3}\left|\begin{array}{ccc}1& 612& 915\\ 101& 2& 3\\ \frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}& \frac{\mathrm{n}(\mathrm{n}+1)}{2}& \mathrm{n}\end{array}\right|\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \left|\begin{array}{ccc}1& 612& 915\\ 101& 2& 3\\ \frac{(\mathrm{n}+1)(2\mathrm{n}+1)}{6{\mathrm{n}}^2}& \frac{\mathrm{n}+1}{2{\mathrm{n}}^2}& \frac{1}{{\mathrm{n}}^2}\end{array}\right|\\ =\left|\begin{array}{ccc}1& 612& 915\\ 101& 2& 3\\ \frac{1}{3}& 0& 0\end{array}\right|\\ =\frac{1}{3}(3\times 612-2\times 915)\\ =2\end{array}$