Step 1: Replace the cable with an equivalent simply supported beam.
A cable under a uniformly distributed load takes up a parabolic shape, and its horizontal pull is linked to the bending moment of an imaginary simply supported beam carrying the same span and same load. For that beam, the maximum bending moment at midspan is \[ M_{max} = \frac{wL^2}{8} = \frac{6 \times 10^2}{8} = 75\,\mathrm{kN\cdot m}. \]
Step 2: Use the cable theorem.
For a cable, the dip acts like the lever arm balancing that same bending moment, so $H \times d = M_{max}$.
Step 3: Solve for the horizontal tension.
\[ H = \frac{M_{max}}{d} = \frac{75}{3} = 25\,\mathrm{kN}. \]
\[ \boxed{25\,\mathrm{kN}} \]