$\left(x^{3}-\frac{1}{x^{2}}\right)^{n}$
$\quad$ General term, $T_{r+1}={ }^{n} C_{r}\left(x^{3}\right)^{n-r} \cdot\left(-\frac{1}{x^{2}}\right)^{r}$
$\quad={ }^{n} C_{r} \cdot x^{(3 n-3 r)} \cdot(-1)^{r} \cdot x^{-2 r}$
$={ }^{n} C_{r} \cdot(-1)^{r} \cdot x^{(3 n-5 r)}$ ...(i)
For the coefficient $\mathrm{x}^{5}$
Put $3 n-5 r=5$
$5 r=3 n-5$
$\therefore \mathrm{r}=\frac{3 \mathrm{n}}{5}-1$
$\therefore$ Coefficient of $x^{5}={ }^{n} C\left(\frac{3 n}{5}-1\right)^{(-1)}\left(\frac{3 n}{5}-1\right)$
For the coefficient of $\mathrm{x}^{10}$
Put $3 n-5 r=10$
$5 r=3 n-10$
$\therefore \mathrm{r}=\frac{3 \mathrm{n}}{5}-2$
$\therefore$ Coefficient of $x^{10}={ }^{n} C_{\left(\frac{3 n}{5}-2\right)^{(-1)}}\left(\frac{3 n}{5}-2\right)$
The sum of the coefficient of $x^{5}$ and $x^{10}=0$
$\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-1\right)^{(-1)}\left(\frac{3 \mathrm{n}}{5}-1\right)}+{ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-2\right)}(-1)^{\left(\frac{3 \mathrm{n}}{5}-2\right)}=0$
$\Rightarrow$
$\left.(-1)^{\frac{3 n}{5}}\left[{ }^{n} C_{\left(\frac{3 n}{5}-1\right)} \cdot(-1)^{-1}+{ }^{n} C_{\left(\frac{3 n}{5}-2\right.}\right) \cdot(-1)^{(-2)}\right]=0$
$\Rightarrow-{ }^{n} C_{\left(\frac{3 n}{5}-1\right)}+{ }^{n} C_{\left(\frac{3 n}{5}-2\right)}=0$ ...(ii)

From equation (ii)
${ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-2\right)}={ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-1\right)}$
$\Rightarrow \mathrm{n}=\left(\frac{3 \mathrm{n}}{5}-2\right)+\left(\frac{3 \mathrm{n}}{5}-1\right)$
$\left[\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{y}} \Rightarrow \mathrm{n}=\mathrm{x}+\mathrm{y}\right]$
$\Rightarrow n=\frac{6 n}{5}-3 \Rightarrow \frac{6 n}{5}-n=3$
$\Rightarrow \frac{\mathrm{n}}{5}=3 \quad \therefore \mathrm{n}=15$