Determine the maximum possible real value of the number $ k$, such that
\[ (a \plus{} b \plus{} c)\left (\frac {1}{a \plus{} b} \plus{} \frac {1}{c \plus{} b} \plus{} \frac {1}{a \plus{} c} \minus{} k \right )\ge k\]
for all real numbers $ a,b,c\ge 0$ with $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca$.
To determine the maximum possible real value of \( k \) such that
\[
(a + b + c) \left( \frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} - k \right) \ge k
\]
for all real numbers \( a, b, c \ge 0 \) with \( a + b + c = ab + bc + ca \), we proceed as follows:

1. **Substitute specific values to find an upper bound for \( k \)**:
   Let \( a = 0 \), \( b = 2 \), and \( c = 2 \). Then,
   \[
   a + b + c = 0 + 2 + 2 = 4
   \]
   and
   \[
   ab + bc + ca = 0 \cdot 2 + 2 \cdot 2 + 2 \cdot 0 = 4
   \]
   Thus, the condition \( a + b + c = ab + bc + ca \) is satisfied.

   Now, we evaluate the left-hand side of the inequality:
   \[
   (a + b + c) \left( \frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} - k \right)
   \]
   Substituting \( a = 0 \), \( b = 2 \), and \( c = 2 \):
   \[
   4 \left( \frac{1}{0 + 2} + \frac{1}{2 + 2} + \frac{1}{2 + 0} - k \right) = 4 \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{2} - k \right)
   \]
   Simplifying inside the parentheses:
   \[
   4 \left( \frac{1}{2} + \frac{1}{2} + \frac{1}{4} - k \right) = 4 \left( 1 + \frac{1}{4} - k \right) = 4 \left( 1.25 - k \right)
   \]
   The inequality becomes:
   \[
   4(1.25 - k) \ge k
   \]
   Simplifying:
   \[
   5 - 4k \ge k
   \]
   \[
   5 \ge 5k
   \]
   \[
   k \le 1
   \]

2. **Verify if \( k = 1 \) satisfies the inequality for all \( a, b, c \ge 0 \)**:
   We need to show that:
   \[
   (a + b + c) \left( \frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} - 1 \right) \ge 1
   \]
   Given \( a + b + c = ab + bc + ca \), we can use the identity:
   \[
   \frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} \ge \frac{9}{2(a + b + c)}
   \]
   This is a known inequality for non-negative \( a, b, c \).

   Therefore:
   \[
   (a + b + c) \left( \frac{9}{2(a + b + c)} - 1 \right) \ge 1
   \]
   Simplifying:
   \[
   (a + b + c) \left( \frac{9 - 2(a + b + c)}{2(a + b + c)} \right) \ge 1
   \]
   \[
   \frac{9 - 2(a + b + c)}{2} \ge 1
   \]
   \[
   9 - 2(a + b + c) \ge 2
   \]
   \[
   9 - 2 \ge 2(a + b + c)
   \]
   \[
   7 \ge 2(a + b + c)
   \]
   \[
   a + b + c \le \frac{7}{2}
   \]
   Since \( a + b + c = ab + bc + ca \), this condition is satisfied for non-negative \( a, b, c \).

Thus, the maximum possible value of \( k \) is \( 1 \).

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