The lines $AD , BE$ and $CF$ intersect in $S$ within a triangle $ABC$ . 
It is given that $AS: DS = 3: 2$ and $BS: ES = 4: 3$ . Determine the ratio $CS: FS$ .

[asy]
unitsize (1 cm);

pair A, B, C, D, E, F, S;

A = (0,0);
B = (5,0);
C = (1,4);
S = (14*A + 15*B + 6*C)/35;
D = extension(A,S,B,C);
E = extension(B,S,C,A);
F = extension(C,S,A,B);

draw(A--B--C--cycle);
draw(A--D);
draw(B--E);
draw(C--F);

dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, N);
dot("$D$", D, NE);
dot("$E$", E, W);
dot("$F$", F, dir(270));
dot("$S$", S, NE);
[/asy]
1. Given the ratios \( \frac{AS}{DS} = \frac{3}{2} \) and \( \frac{BS}{ES} = \frac{4}{3} \), we need to determine the ratio \( \frac{CS}{FS} \).

2. First, let's express the ratios in terms of the segments of the triangle:
   - Since \( \frac{AS}{DS} = \frac{3}{2} \), we can write \( AS = 3k \) and \( DS = 2k \) for some constant \( k \). Therefore, \( AD = AS + DS = 3k + 2k = 5k \).
   - Similarly, since \( \frac{BS}{ES} = \frac{4}{3} \), we can write \( BS = 4m \) and \( ES = 3m \) for some constant \( m \). Therefore, \( BE = BS + ES = 4m + 3m = 7m \).

3. Using the area ratios, we can find the areas of the triangles formed by the cevians:
   - The area of \( \triangle ASB \) is proportional to \( \frac{AS}{AD} = \frac{3k}{5k} = \frac{3}{5} \).
   - The area of \( \triangle BSD \) is proportional to \( \frac{DS}{AD} = \frac{2k}{5k} = \frac{2}{5} \).

4. Similarly, for the other cevian:
   - The area of \( \triangle BSC \) is proportional to \( \frac{BS}{BE} = \frac{4m}{7m} = \frac{4}{7} \).
   - The area of \( \triangle ESC \) is proportional to \( \frac{ES}{BE} = \frac{3m}{7m} = \frac{3}{7} \).

5. Now, we need to find the ratio \( \frac{CS}{FS} \). Let \( CS = x \) and \( FS = y \). Then \( CF = CS + FS = x + y \).

6. Using the area ratios, we can write:
   - The area of \( \triangle ASC \) is proportional to \( \frac{AS}{AD} = \frac{3}{5} \).
   - The area of \( \triangle DSC \) is proportional to \( \frac{DS}{AD} = \frac{2}{5} \).

7. Combining the ratios, we get:
   - The area of \( \triangle ASC \) is proportional to \( \frac{3}{5} \times \frac{4}{7} = \frac{12}{35} \).
   - The area of \( \triangle DSC \) is proportional to \( \frac{2}{5} \times \frac{3}{7} = \frac{6}{35} \).

8. Therefore, the ratio of the areas of \( \triangle ASC \) to \( \triangle DSC \) is \( \frac{12}{35} : \frac{6}{35} = 2:1 \).

9. Since the areas are proportional to the lengths of the segments, we have \( \frac{CS}{FS} = 2:1 \).

The final answer is \( \boxed{2:1} \).