Let P(3secθ,2tanθ) and Q(3secφ,2tanφ), where θ+φ=π/2, be two distinct points on the hyperbola x²/9 - y²/4 = 1. Then the ordinate of the point of intersection of the normals at P and Q is:

Equation of normal is, a²x/x₁ + b²y/y₁ = a² + b²
The equation of the hyperbola is x²/9 - y²/4 = 1. Here a² = 9 and b² = 4.
Let P(x₁, y₁) = (3secθ, 2tanθ) and Q(x₂, y₂) = (3secφ, 2tanφ).
The equation of the normal at P is:
9x/(3secθ) + 4y/(2tanθ) = 9 + 4
3x/secθ + 2y/tanθ = 13
3xcosθ + 2ycotθ = 13
The equation of the normal at Q is:
9x/(3secφ) + 4y/(2tanφ) = 9 + 4
3x/secφ + 2y/tanφ = 13
3xcosφ + 2ycotφ = 13
Since θ + φ = π/2, we have φ = π/2 - θ.
Substituting this into the equation of the normal at Q:
3xcos(π/2 - θ) + 2ycot(π/2 - θ) = 13
3xsinθ + 2ytanθ = 13
Now we have a system of two equations:

1. 3xcosθ + 2ycotθ = 13
2. 3xsinθ + 2ytanθ = 13
   Multiply equation (1) by sinθ and equation (2) by cosθ:
   3xsinθcosθ + 2ysinθcotθ = 13sinθ
   3xsinθcosθ + 2ycosθtanθ = 13cosθ
   Subtracting the two equations:
   2y(sinθcotθ - cosθtanθ) = 13(sinθ - cosθ)
   2y(sinθ(cosθ/sinθ) - cosθ(sinθ/cosθ)) = 13(sinθ - cosθ)
   2y(cosθ - sinθ) = 13(sinθ - cosθ)
   2y = -13
   y = -13/2
   Therefore, the ordinate of the point of intersection of the normals at P and Q is -13/2 = -6.5. However, this value is not among the options. There might be a mistake in the problem statement or the given options.