Let $ a \in \mathbb{R}, 0 < a < 1,$ and $ f$ a continuous function on $ [0, 1]$ satisfying $ f(0) \equal{} 0, f(1) \equal{} 1,$ and

\[ f \left( \frac{x\plus{}y}{2} \right) \equal{} (1\minus{}a) f(x) \plus{} a f(y) \quad \forall x,y \in [0,1] \text{ with } x \leq y.\]

Determine $ f \left( \frac{1}{7} \right).$
1. Given the function \( f \) is continuous on \([0, 1]\) and satisfies the conditions \( f(0) = 0 \), \( f(1) = 1 \), and 
   \[
   f \left( \frac{x+y}{2} \right) = (1-a) f(x) + a f(y) \quad \forall x,y \in [0,1] \text{ with } x \leq y.
   \]
   We need to determine \( f \left( \frac{1}{7} \right) \).

2. Let's consider the functional equation:
   \[
   f \left( \frac{x+y}{2} \right) = (1-a) f(x) + a f(y).
   \]
   We will use this equation iteratively to find a pattern.

3. First, set \( x = 0 \) and \( y = 1 \):
   \[
   f \left( \frac{0+1}{2} \right) = f \left( \frac{1}{2} \right) = (1-a) f(0) + a f(1).
   \]
   Given \( f(0) = 0 \) and \( f(1) = 1 \), we have:
   \[
   f \left( \frac{1}{2} \right) = (1-a) \cdot 0 + a \cdot 1 = a.
   \]

4. Next, set \( x = 0 \) and \( y = \frac{1}{2} \):
   \[
   f \left( \frac{0 + \frac{1}{2}}{2} \right) = f \left( \frac{1}{4} \right) = (1-a) f(0) + a f \left( \frac{1}{2} \right).
   \]
   Using \( f(0) = 0 \) and \( f \left( \frac{1}{2} \right) = a \), we get:
   \[
   f \left( \frac{1}{4} \right) = (1-a) \cdot 0 + a \cdot a = a^2.
   \]

5. Now, set \( x = 0 \) and \( y = \frac{1}{4} \):
   \[
   f \left( \frac{0 + \frac{1}{4}}{2} \right) = f \left( \frac{1}{8} \right) = (1-a) f(0) + a f \left( \frac{1}{4} \right).
   \]
   Using \( f(0) = 0 \) and \( f \left( \frac{1}{4} \right) = a^2 \), we get:
   \[
   f \left( \frac{1}{8} \right) = (1-a) \cdot 0 + a \cdot a^2 = a^3.
   \]

6. Continuing this pattern, we can generalize that:
   \[
   f \left( \frac{1}{2^n} \right) = a^n.
   \]

7. To find \( f \left( \frac{1}{7} \right) \), we need to express \(\frac{1}{7}\) in terms of the form \(\frac{1}{2^n}\). However, since \(\frac{1}{7}\) is not of the form \(\frac{1}{2^n}\), we need to use the continuity of \( f \) and the functional equation to approximate it.

8. By the continuity of \( f \) and the functional equation, we can infer that \( f \left( \frac{1}{7} \right) \) will be a value between \( f \left( \frac{1}{8} \right) = a^3 \) and \( f \left( \frac{1}{4} \right) = a^2 \). Given the pattern, we can approximate \( f \left( \frac{1}{7} \right) \) as a value close to \( a^3 \).

The final answer is \( \boxed{ f \left( \frac{1}{7} \right) \approx a^3 } \).