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

$$\text{The given vectors } \overrightarrow{AB}=B-A\text{ and }\overrightarrow{AC}=C-A \text{ and the solution }D=A+\overrightarrow{AD}$$

Let $(x,y)$ be the coordinates of $D$. The equation
$$5\overrightarrow{AD}=2\overrightarrow{AB}-3\overrightarrow{AC}\tag{0}$$
means that
$$5(x-4,y-2)=2(-2-4,1-2)-3(3-4,-2-2),\tag{1}$$
because the vectors $\overrightarrow{AD}=D-A$, $\overrightarrow{AB}=B-A$ and 
$\overrightarrow{AC}=C-A$.
The vectors $5\overrightarrow{AD}=5\left( D-A\right) =\left( 5D-5A\right) $, 
$2\overrightarrow{AB}=\left( 2B-2A\right) $, etc.
A possible way of solving the equation $(1)$ is as follows.
$$5(x-4,y-2)=2(-2-4,1-2)-3(3-4,-2-2)$$
$$\begin{eqnarray*}
&\Leftrightarrow &(5x-20,5y-10)=2(-6,-1)-3(-1,-4) \\
&\Leftrightarrow &(5x-20,5y-10)=(-12,-2)-(-3,-12) \\
&\Leftrightarrow &(5x-20,5y-10)=(-12,-2)+(3,12) \\
&\Leftrightarrow &(5x-20,5y-10)=(-12+3,-2+12) \\
&\Leftrightarrow &(5x-20,5y-10)=(-9,10) \\
&\Leftrightarrow &\left\{ 
\begin{array}{c}
5x-20=-9 \\ 
5y-10=10
\end{array}
\right. \Leftrightarrow \left\{ 
\begin{array}{c}
x=\frac{11}{5} \\ 
y=4
\end{array}
\right. 
\end{eqnarray*}\tag{2}$$
