So I have the image $\displaystyle \operatorname{Im}f=\{\lambda_1(1, 2 ,0)+\lambda_2(2 ,1, 3) \}$ and I have to find the values of $\lambda$ so that the vector $\displaystyle (1,\lambda,\lambda^{2}) \in \operatorname{Im}(f)$.
Can you tell me just where to start? Just a little help?
You should solve the following equation $(1,\lambda, \lambda^{2})-\lambda_{1}(1,2,0)-\lambda_{2}(2,1,3)=(0,0,0)$. Which is in fact a system of three following equations: $$ 1-\lambda_{1}-2\lambda_{2}=0, \\ \lambda -2\lambda_1-\lambda_2 = 0,\\ \lambda^{2}- 3\lambda_{2}=0.$$
The generic vector of the image is $(v_1,v_2,v_3)\equiv(\lambda_1+2\lambda_2, 2\lambda_1+\lambda_2,3\lambda_2)$. So you need a $\lambda$ such that $v_1=1$, $v_2=\lambda$, $v_3=\lambda^2$. This is a system of equations.
Playing with it, you should find a quadratic condition on $\lambda$, namely $\lambda^2+\lambda-2=0$.
You should have seen that the values of $\lambda$ you are looking for are the roots of the polynomial in $\lambda$ given by $$ \det \begin{bmatrix} 1&2&0\\ 2&1&3\\ 1&\lambda&\lambda^2 \end{bmatrix}. $$
