PQR is a triangle. $\overrightarrow{PQ}=\begin{pmatrix}2\\-3\end{pmatrix}$, $\overrightarrow{PR}=\begin{pmatrix}-3\\9\end{pmatrix}$ and Q is the point (0, 5).

1. Use $\overrightarrow{PQ}$ and the coordinates of Q to find P

Let $P=(x,y)$. Since $Q=(0,5)$ and $\overrightarrow{PQ}=\begin{pmatrix}2\\-3\end{pmatrix}$,

$\begin{pmatrix}0-x\\5-y\end{pmatrix}=\begin{pmatrix}2\\-3\end{pmatrix}$

So $0-x=2$ and $5-y=-3$.

Hence $x=-2$ and $y=8$.

1. Find the vector from Q to R

$\overrightarrow{QR}=\overrightarrow{PR}-\overrightarrow{PQ}$
$=\begin{pmatrix}-3\\9\end{pmatrix}-\begin{pmatrix}2\\-3\end{pmatrix}$
$=\begin{pmatrix}-5\\12\end{pmatrix}$

2. Calculate the length of QR from its vector components

$QR=\sqrt{(-5)^2+12^2}$
$=\sqrt{25+144}$
$=\sqrt{169}=13$