This problem has been troubling me.
Let $$P(x) = x^3 − 2x^2 − 11x + d \quad (a \in \mathbb{R} )$$ The roots of $P(x) = 0$ are $p, q, r$ and $p + 2q + 3r = 0$. Find all possible values of $d$.
I used Vieta's formulas to set up three equations $$\begin{align*}2 & = p + q + r \\ -11 &= pq + pr + qr \\ -d &= pqr \end{align*}$$
I tried solving for $d$, but made no progress. I know how to solve the system if there were three variables, but the extra variable $d$ probably requires some algebraic manipulation to find. So how do I find $d$?
$$\begin{align*}2 & = p + q + r \\ -11 &= pq + pr + qr \\ -d &= pqr \\0 &= p+2q+3r\end{align*}$$ From the first and last equations you can get $$\begin{align*}p&=2-q-r\\0&=2-q-r+2q+3r=2+q+2r\end{align*}$$ So if we substitute back: $$\begin{align*}q&=-2-2r\\p&=2+2+2r-r=4+r\end{align*}$$ You can now plug these into the second equation in the original set. You should get a quadratic in $r$. You solve it, find $p$ and $q$, and then use those to find $d$