3 known, 3 unknown

This solution use the theorem 'ratio of length = ratio area of triangle with same height' frequently (Exp: $\frac {CF}{FP}= \frac{Area \triangle CFB}{Area \triangle PFB}$ ). Apply Ceva theorem with $\triangle BPC$ and point $A$ . We get $\begin{aligned} &\frac{BD}{DC}× \frac{CF}{FP} × \frac{PE}{EB}=1 \\ & \frac{2}{3}× \frac{1+2+3}{1}× \frac{c}{c+3+2}=1 \\ & 4c=c+5 \\ & c= \frac {5}{3} \end{aligned}$ Apply Ceva theorem with $\triangle APB$ and point $C$ . We get $\begin{aligned} & \frac{AF}{FB}× \frac{BE}{EP}× \frac{PD}{DA}=1 \\ & \frac{a}{1}× \frac{2+3+c}{c}× \frac{2}{2+1+a}=1 \\ & 8a=a+3 \\ & a= \frac{3}{7} \end{aligned}$ Apply Ceva theorem with $\triangle ABC$ and point $P$ . We get $\begin{aligned} & \frac{AF}{FB}× \frac{BD}{DC}× \frac{CE}{EA}=1 \\ & \frac{a}{1}× \frac{2}{3}× \frac{c}{b}=1 \\ & b= \frac{10}{21} \end{aligned}$ Hence, the answer is $7a+21b+9c=\large 28$ .

The key idea is that if 2 triangles have the same height then the ratio of their areas is equal to the ratio of there base lengths. $\frac{[ABD]}{[ACD]} = \frac{BD}{CD}$ But we also have $\frac{[PBD]}{[PCD]} = \frac{BD}{CD}$

Thus $\frac{[ABD]}{[ACD]} = \frac{[PBD]}{[PCD]}$

$\frac{a+3}{b+c+3} = \frac{2}{3}$

$3a+3=2b+2c$

Working similarly with other pairs of triangles gives $5a=b+c$ and $c+ca=5b$

Plugging $5a=b+c$ into $3a+3=2b+2c$ gives $a=3/7$

From there we can get $c=5/3$ and $b=10/21$