Refer to figure.
Consider an aircraft in a straight steady climb along a straight flight path inclined at an angle (γ) to the horizontal. γ (gamma) is the symbol used for climb angle. The forces on the aircraft consist of Lift, normal to the flight path; Thrust and Drag, parallel to it; and Weight, parallel to the force of gravity.

Weight is resolved into two components: one opposite Lift (W cos γ) and the other acting in the same direction as Drag (W sin γ), backwards along the flight path. The requirements for equilibrium are: Thrust must equal the sum of Drag plus the backwards component of Weight; and Lift must equal its opposing component of Weight. For equilibrium at a greater angle of climb, the Lift required will be less, and the backwards component of Weight will be greater.

L = W cosγ
T = D + W sinγ

In a straight steady climb, Lift is less than Weight because Lift only has to support a proportion of the weight, this proportion decreasing as the climb angle increases. The remaining proportion of Weight is supported by engine Thrust.
It can be seen that for a straight steady climb the Thrust required is greater than Drag. This is to balance the backward component of Weight acting along the flight path.

The climb gradient can be described as:

sin γ = (T−D)/W

For very shallow flight path angles (γ), we can assume that lift is approximately equal to Weight, therefore the climb gradient can be described as:

sin γ = T/W−D/L

In this exercise, Weight is equal to 500000 N, D/L ratio is equal to 1/12 and Thrust equal to 28000 N per engine.

For the all engines (three) climb gradient and applying the formula defined above:

sin γ = (84000/500000) −(1/12)
sinγ = 0.085 ⇒ sinγ = 8.5%

The climb gradient is 8.5%.