Find the dual basis $(w_1,w_2,w_3)$ of $\mathbb R^3$, meaning that $$\langle v_i,w_j\rangle = \begin{cases} 1 \mbox{ if } i=j, \\ 0 \mbox{ if } i \ne j ,\end{cases}$$ for $i,j = 1,2,3, $ where $$v_1= (1,2,0), \ \ \ \ \ v_2 = (2,3,1), \ \ \ \ \ v_3 = (-1,-1,2).$$
How would I even start a question like this?
If $\{ v_1, v_2, v_3 \}$ is a basis of $\mathbb R^3$, then the vectors $$ w_1 = \frac{v_2 \times v_3}{v_1.(v_2 \times v_3)}, \ \ \ \ \ w_2 = \frac{v_3 \times v_1}{v_2.(v_3 \times v_1)}, \ \ \ \ \ \ \ \ \ \ w_3 = \frac{v_1 \times v_2}{v_3.(v_1 \times v_2)}$$ obey the conditions $$ v_i . w_j = \begin{cases} 1 & i = j \\ 0 & i \neq j\end{cases}$$ [Just calculate the dot products and you'll see!]
And $\{w_1, w_2, w_3 \}$ form a basis of $\mathbb R^3$ because their scalar triple product is non-zero: $$ w_1 . (w_2 \times w_3) = \frac 1 {v_1 . (v_2 \times v_3)} \neq 0.$$ [I encourage you to verify this from the identities for scalar and vector triple products - see here.]