Problem : Let $ABC$ triangle , $A',B',C'$ point respectively in $[BC],[AC],[BA]$ Such that :
$\vec{CB}=3\vec{CA'}$
$\vec{BA}=3\vec{BC'}$
$\vec{AC}=3\vec{AB'}$
Line $(CC')$ insert with $(BB')$ at $R$
Line $(AA')$ insert with $(BB')$ at $T$
Line $(CC')$ insert with $(AA')$ at $S$
Now : $kS_{RST}=S_{ABC}$ , $S=$ area
Then find the value of $k$ ?
I need help to fine this value of $k$ I don't know which Theorem must be use it
I have already to see your hints and ideas
Here's a fun solution using Barycentric coordinates. In Barycentric coordinates, you express points $P$ in the plane as a normalized linear combination of a triangle ABC's vertices, i.e. if $P = xA+yB+zC$ with $x+y+z=1$ then $P$ is represented as $(x,y,z)$. A great reference is https://s3.amazonaws.com/aops-cdn.artofproblemsolving.com/resources/articles/bary.pdf.
Since if we view points as vectors, $B' = A + \frac{1}{3}(C-A) = \frac{2}{3}A + \frac{1}{3}C$, we get $B' = (\frac{2}{3},0,\frac{2}{3})$ and similarly that $A' = (0,\frac{1}{3},\frac{2}{3})$ and $C' = (\frac{1}{3},\frac{2}{3},0)$. Next, we compute the equations for the lines $AA'$, $BB'$ and $CC'$, and then compute their intersections $R, S, T$.
We compute $AA'$ as follows: lines through vertex $A$ have the form $y=kz$ for some $k$, and since $A'$ is on this line, we have that $k=\frac{1}{2}$. Thus (and similarly for $BB'$, $CC'$), we get that $AA'$ has equation $y=\frac{1}{2}z$, $BB'$ has equation $z=\frac{1}{2}x$ and $CC'$ has equation $x=\frac{1}{2}y$.
Since $T$ is on $AA', BB'$, its coordinates are $(\frac{4}{7},\frac{1}{7},\frac{2}{7})$. Similarly, $S$ has coordinates $(\frac{1}{7},\frac{2}{7},\frac{4}{7})$ and $R$ has coordinates $(\frac{2}{7},\frac{4}{7},\frac{1}{7})$.
Now, we use the fact that if $R,S,T$ have coordinates $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)$, then the area of $RST$ divided by the area of $ABC$ is given by the absolute value of the determinant \begin{vmatrix}x_1&y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3\end{vmatrix} We have that $$\begin{vmatrix}\frac{4}{7}&\frac{1}{7}&\frac{2}{7}\\ \frac{2}{7}&\frac{4}{7}&\frac{1}{7}\\ \frac{1}{7}&\frac{2}{7}&\frac{4}{7}\end{vmatrix} = \frac{1}{7}$$ And so $k=7$ as desired.