${\mathrm{\Delta }}_r=\left|\begin{array}{ccc}r& \left(2r-1\right)& \left(3r-2\right)\\ \frac{n}{2}& \left(n-1\right)& a\\ \frac{1}{2}n\left(n-1\right)& {\left(n-1\right)}^2& \frac{1}{2}\left(n-1\right)\left(3n+4\right)\end{array}\right|$

Since, the second and the third rows are independent of $r,$ hence the sum is applied to the first row only.

$\Rightarrow \sum _{r=1}^{n-1}{\mathrm{\Delta }}_r=\left|\begin{array}{ccc}\sum _{r=1}^{n-1}r& 2\sum _{r=1}^{n-1}r-\sum _{r=1}^{n-1}1& 3\sum _{r=1}^{n-1}r-2\sum _{r=1}^{n-1}1\\ \frac{n}{2}& \left(n-1\right)& a\\ \frac{1}{2}n\left(n-1\right)& {\left(n-1\right)}^2& \frac{1}{2}\left(n-1\right)\left(3n+4\right)\end{array}\right|$

Using $\sum _{r=1}^{n}r=\frac{n\left(n+1\right)}{2},$ we get

$\sum _{\mathrm{r}=1}^{\mathrm{n}-1}{\mathrm{\Delta }}_{\mathrm{r}}\mathrm{=}\left|\begin{array}{ccc}\frac{\left(n-1\right)n}{2}& \frac{2\left(n-1\right)n}{2}-\left(n-1\right)& \frac{3\left(n-1\right)n}{2}-2\left(n-1\right)\\ \frac{n}{2}& \left(n-1\right)& a\\ \frac{1}{2}n\left(n-1\right)& {\left(n-1\right)}^2& \frac{1}{2}\left(n-1\right)\left(3n+4\right)\end{array}\right|$

$\Rightarrow \sum _{\mathrm{r}=1}^{\mathrm{n}-1}{\mathrm{\Delta }}_{\mathrm{r}}\mathrm{=}\left|\begin{array}{ccc}\frac{1}{2}n\left(n-1\right) & {\left(n-1\right)}^2& \frac{1}{ 2}\left(n-1\right)\left(3n+4\right)\\ \frac{n}{2}& \left(n-1\right)& a\\ \frac{1}{2}n\left(n-1\right) & {\left(n-1\right)}^2& \frac{1}{2}\left(n-1\right)\left(3n+4\right)\end{array}\right|$

$\Rightarrow \sum _{r=1}^{n-1}{\mathrm{\Delta }}_r=0,$ $(\because R_1$ and $R_3$ are identical$)$

Hence, the sum is independent of both $n$ and $a.$