Let $S$ be the set of positive real numbers. Find all functions $f\colon S^3 \to S$ such that, for all positive real numbers $x$, $y$, $z$ and $k$, the following three conditions are satisfied:

(a) $xf(x,y,z) = zf(z,y,x)$,

(b) $f(x, ky, k^2z) = kf(x,y,z)$,

(c) $f(1, k, k+1) = k+1$.

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1. **Using Condition (c):**
   \[
   f(1, k, k+1) = k+1
   \]
   This gives us a specific value for \( f \) when the first argument is 1.

2. **Using Condition (b):**
   \[
   f(x, ky, k^2z) = kf(x,y,z)
   \]
   Let's set \( x = 1 \), \( y = k \), and \( z = k+1 \):
   \[
   f(1, k, k^2(k+1)) = kf(1, k, k+1)
   \]
   From Condition (c), we know \( f(1, k, k+1) = k+1 \):
   \[
   f(1, k, k^2(k+1)) = k(k+1)
   \]
   This implies:
   \[
   f(1, k, k^2(k+1)) = k(k+1)
   \]

3. **Using Condition (a):**
   \[
   xf(x,y,z) = zf(z,y,x)
   \]
   Let's set \( x = 1 \), \( z = k+1 \):
   \[
   f(1, y, k+1) = \frac{k+1}{1} f(k+1, y, 1)
   \]
   From Condition (c), we know \( f(1, k, k+1) = k+1 \):
   \[
   k+1 = (k+1) f(k+1, y, 1)
   \]
   This implies:
   \[
   f(k+1, y, 1) = 1
   \]

4. **General Form of \( f \):**
   Given the homogeneity and symmetry conditions, we can guess a form for \( f \). Let's assume:
   \[
   f(x, y, z) = \frac{y + \sqrt{y^2 + 4xz}}{2x}
   \]
   We need to verify if this form satisfies all the conditions.

   - **Condition (a):**
     \[
     xf(x,y,z) = x \cdot \frac{y + \sqrt{y^2 + 4xz}}{2x} = \frac{y + \sqrt{y^2 + 4xz}}{2}
     \]
     \[
     zf(z,y,x) = z \cdot \frac{y + \sqrt{y^2 + 4zx}}{2z} = \frac{y + \sqrt{y^2 + 4zx}}{2}
     \]
     Since \( \sqrt{y^2 + 4xz} = \sqrt{y^2 + 4zx} \), Condition (a) is satisfied.

   - **Condition (b):**
     \[
     f(x, ky, k^2z) = \frac{ky + \sqrt{(ky)^2 + 4x(k^2z)}}{2x} = \frac{ky + \sqrt{k^2y^2 + 4kx^2z}}{2x}
     \]
     \[
     kf(x,y,z) = k \cdot \frac{y + \sqrt{y^2 + 4xz}}{2x} = \frac{ky + k\sqrt{y^2 + 4xz}}{2x}
     \]
     Since \( \sqrt{k^2(y^2 + 4xz)} = k\sqrt{y^2 + 4xz} \), Condition (b) is satisfied.

   - **Condition (c):**
     \[
     f(1, k, k+1) = \frac{k + \sqrt{k^2 + 4(1)(k+1)}}{2} = \frac{k + \sqrt{k^2 + 4k + 4}}{2} = \frac{k + (k+2)}{2} = k+1
     \]
     Condition (c) is satisfied.

Thus, the function \( f(x, y, z) = \frac{y + \sqrt{y^2 + 4xz}}{2x} \) satisfies all the given conditions.

The final answer is \( \boxed{\frac{y + \sqrt{y^2 + 4xz}}{2x}} \).