Given expression is $\left(1-x+2 x^3\right)^{10}$

So, general term is $\frac{10 !}{r_{1} ! r_{2} ! r_{3} !}(1)^{r_1}(-1)^{r_2} \cdot(2)^{r_3} \cdot(x)^{r_2+r_3}$

Where, $r_1+r_2+r_3=10$ and $r_2+3 r_3=7$

Now, for possibility,

$\begin{array}{ccc}r_1 & r_2 & r_3 \\ 3 & 7 & 0 \\ 7 & 1 & 2 \\ 5 & 4 & 1\end{array}$

Thus, required co-efficient