Let $ABC$ be an isosceles triangle with $AB=4$, $BC=CA=6$. On the segment $AB$ consecutively lie points $X_{1},X_{2},X_{3},\ldots$ such that the lengths of the segments $AX_{1},X_{1}X_{2},X_{2}X_{3},\ldots$ form an infinite geometric progression with starting value $3$ and common ratio $\frac{1}{4}$. On the segment $CB$ consecutively lie points $Y_{1},Y_{2},Y_{3},\ldots$ such that the lengths of the segments $CY_{1},Y_{1}Y_{2},Y_{2}Y_{3},\ldots$ form an infinite geometric progression with starting value $3$ and common ratio $\frac{1}{2}$. On the segment $AC$ consecutively lie points $Z_{1},Z_{2},Z_{3},\ldots$ such that the lengths of the segments $AZ_{1},Z_{1}Z_{2},Z_{2}Z_{3},\ldots$ form an infinite geometric progression with starting value $3$ and common ratio $\frac{1}{2}$. Find all triplets of positive integers $(a,b,c)$ such that the segments $AY_{a}$, $BZ_{b}$ and $CX_{c}$ are concurrent.
1. **Identify the geometric progressions on each segment:**
   - On segment \( AB \):
     \[
     AX_1 = 3, \quad X_1X_2 = 3 \cdot \frac{1}{4} = \frac{3}{4}, \quad X_2X_3 = \frac{3}{4} \cdot \frac{1}{4} = \frac{3}{16}, \quad \text{and so on.}
     \]
     The sum of this infinite geometric series is:
     \[
     S = \frac{3}{1 - \frac{1}{4}} = \frac{3}{\frac{3}{4}} = 4.
     \]
   - On segment \( CB \):
     \[
     CY_1 = 3, \quad Y_1Y_2 = 3 \cdot \frac{1}{2} = \frac{3}{2}, \quad Y_2Y_3 = \frac{3}{2} \cdot \frac{1}{2} = \frac{3}{4}, \quad \text{and so on.}
     \]
     The sum of this infinite geometric series is:
     \[
     S = \frac{3}{1 - \frac{1}{2}} = \frac{3}{\frac{1}{2}} = 6.
     \]
   - On segment \( AC \):
     \[
     AZ_1 = 3, \quad Z_1Z_2 = 3 \cdot \frac{1}{2} = \frac{3}{2}, \quad Z_2Z_3 = \frac{3}{2} \cdot \frac{1}{2} = \frac{3}{4}, \quad \text{and so on.}
     \]
     The sum of this infinite geometric series is:
     \[
     S = \frac{3}{1 - \frac{1}{2}} = \frac{3}{\frac{1}{2}} = 6.
     \]

2. **Determine the positions of \( X_c \), \( Y_a \), and \( Z_b \):**
   - For \( X_c \) on \( AB \):
     \[
     AX_c = 3 \left(1 + \frac{1}{4} + \left(\frac{1}{4}\right)^2 + \cdots + \left(\frac{1}{4}\right)^{c-1}\right) = 3 \cdot \frac{1 - \left(\frac{1}{4}\right)^c}{1 - \frac{1}{4}} = 4 \left(1 - \left(\frac{1}{4}\right)^c\right).
     \]
   - For \( Y_a \) on \( CB \):
     \[
     CY_a = 3 \left(1 + \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \cdots + \left(\frac{1}{2}\right)^{a-1}\right) = 3 \cdot \frac{1 - \left(\frac{1}{2}\right)^a}{1 - \frac{1}{2}} = 6 \left(1 - \left(\frac{1}{2}\right)^a\right).
     \]
   - For \( Z_b \) on \( AC \):
     \[
     AZ_b = 3 \left(1 + \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \cdots + \left(\frac{1}{2}\right)^{b-1}\right) = 3 \cdot \frac{1 - \left(\frac{1}{2}\right)^b}{1 - \frac{1}{2}} = 6 \left(1 - \left(\frac{1}{2}\right)^b\right).
     \]

3. **Set up the concurrency condition:**
   - The segments \( AY_a \), \( BZ_b \), and \( CX_c \) are concurrent if:
     \[
     4^c - 1 = (2^a - 1)(2^b - 1).
     \]

4. **Apply the lemma:**
   - Using the lemma \( 2^k - 1 \mid 2^\ell - 1 \implies k \mid \ell \):
     \[
     (2^a - 1)(2^b - 1) = 2^{2c} - 1.
     \]
     This implies \( a \mid 2c \) and \( b \mid 2c \).

5. **Find the solutions:**
   - If \( a = 2c \) or \( b = 2c \), we get:
     \[
     (a, b, c) = (1, 2c, c) \quad \text{or} \quad (2c, 1, c).
     \]
   - If \( a \neq 2c \neq b \), we reach a contradiction:
     \[
     2^{2c} - 1 = (2^a - 1)(2^b - 1) \leq (2^{c} - 1)(2^{c} - 1) < (2^c - 1)(2^c + 1) = 2^{2c} - 1.
     \]

The final answer is \( \boxed{ (a, b, c) = (1, 2c, c) \text{ or } (2c, 1, c) } \).