$M$ and $M'$ are points lying on line segment $BC$ of $\triangle ABC$ such that $BM = 2MC$ and $BM' = 3M'C$. $N$ and $N'$ are points lying repsectively on $AM$ and $AM'$ such that $AN = 2AM$ and $AN' = \dfrac{2}{3}AM'$.
a/ Determine $x$ and $y$ such that $\overrightarrow{NN'} = x\overrightarrow{AB} + y\overrightarrow{AC}$.
b/ $NN' \cap BC = P$. Determine the value of $z$ such that $BP = zPC$ and calculate the value of $\dfrac{NP}{PN'}$.
We have that $BM = 2MC$ and $BM' = 3M'C \implies \overrightarrow{CM} = \dfrac{1}{3}\overrightarrow{CB}$ and $\overrightarrow{CM'} = \dfrac{1}{4}\overrightarrow{CB}$.
$\overrightarrow{NN'} = \overrightarrow{AN'} - \overrightarrow{AN} = \dfrac{2}{3}\overrightarrow{AM'} - 2\overrightarrow{AM} = \dfrac{2}{3} \cdot \left(\overrightarrow{AC} + \overrightarrow{CM'}\right) - 2 \cdot \left(\overrightarrow{AC} + \overrightarrow{CM}\right)$
$ = \dfrac{2}{3} \cdot \dfrac{1}{4}\overrightarrow{CB} - 2 \cdot \dfrac{1}{3}\overrightarrow{CB} - \dfrac{4}{3}\overrightarrow{AC} = -\dfrac{1}{2}\overrightarrow{CB} - \dfrac{4}{3}\overrightarrow{AC}$
$ = -\dfrac{1}{2}\left(\overrightarrow{AB} - \overrightarrow{AC}\right) - \dfrac{4}{3}\overrightarrow{AC} = -\dfrac{1}{2}\overrightarrow{AB} - \dfrac{5}{6}\overrightarrow{AC}$
For b/, I am not sure where to start.
a) Trace the vector $\vec{NN'} $ as follows
\begin{align} \vec{NN'} &= \vec{AN'} -\vec{AN} = \frac23 \vec{AM'} - 2 \vec{AM}\\ &= \frac23 (\vec{BM'} - \vec{BA} ) - 2(\vec{BM} - \vec{BA} ) = \frac23 \vec{BM'} -2\vec{BM} +\frac 43\vec{BA} \\ &= \frac23\cdot \frac 34 \vec{BC} -2\cdot\frac 23\vec{BC} +\frac 43\vec{BA} = -\frac56 (\vec{AC} - \vec{AB} )- \frac 43\vec{AB} \\ &= -\frac12 \vec{AB} -\frac56 \vec{AC} \end{align}
Thus
$$x=-\frac12,\>\>\>\>\> y=-\frac56 $$
b) Let [.] denote areas.
$$\frac{MP}{PM'} = \frac{[MNN']}{[NM'N']} = \frac{[MNN']}{\frac 12[NAN']} =\frac{[MNN']}{\frac 12\cdot 2[NM'N']}=1$$
So, $P$ is midpoint of $MM'$. Then,
$$\frac{BP}{PC} = \frac{BM+MP}{PM'+M'C}=\frac{(\frac23+\frac{1}{24})BC}{(\frac{1}{24}+\frac14)BC}=\frac{17}{7}$$
Thus, $$z=\frac{17}{7}$$
Finally,
$$\frac{NP}{PN'} = \frac{[MM'N]}{[MM'N']} = \frac{[MM'N]}{\frac13 [MM'A]} = 3$$