Determine a point $$\text{ABC- triangle:} A(4,2); B(-2,1);C(3,-2)$$
Find a D point so this equality is true:
$$5\vec{AD}=2\vec{AB}-3\vec{AC}$$

Recall that the vector $\overrightarrow{PQ}$ is the difference of two points $Q{-}P$. In this way,
$$
5\overrightarrow{AD}=2\overrightarrow{AB}-3\overrightarrow{AC}
$$
becomes
$$
5(D-A)=2(B-A)-3(C-A)
$$
All that is left is to solve for $D$.
