To solve for the value of $$\alpha$$, which represents the square of the speed of the particle at the time its $$x$$-coordinate is $$84 \mathrm{m}$$, we need to first determine the time at which the particle reaches this $$x$$-coordinate, and then use this time to calculate its final velocity in both the $$x$$ and $$y$$ directions.

The motion of the particle in the $$x$$-direction can be described by the kinematic equation for uniformly accelerated motion:

$$ x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2 $$

Given:

$$ x_0 = 0 \mathrm{~m} $$

$$ v_{0x} = 5 \mathrm{~m/s} $$

$$ a_x = 3 \mathrm{~m/s}^2 $$

$$ x = 84 \mathrm{~m} $$ (at which we need to find the speed)

Substituting these values into the kinematic equation:

$$ 84 \mathrm{~m} = 0 \mathrm{~m} + (5 \mathrm{~m/s})t + \frac{1}{2}(3 \mathrm{~m/s}^2)t^2 $$

Simplifying this equation, we get:

$$ 0 = \frac{3}{2}t^2 + 5t - 84 $$

Using the quadratic formula to solve for $$t$$:

$$ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

where $$a = \frac{3}{2}, b = 5, $$ and $$c = -84$$.

$$ t = \frac{-5 \pm \sqrt{(5)^2 - 4(\frac{3}{2})(-84)}}{2(\frac{3}{2})} $$

$$ t = \frac{-5 \pm \sqrt{25 + 504}}{3} $$

$$ t = \frac{-5 \pm \sqrt{529}}{3} $$

$$ t = \frac{-5 \pm 23}{3} $$

In this scenario, since we're looking for a time when the particle reaches $$84 \mathrm{m}$$, we only consider the positive root because time cannot be negative.

$$ t = \frac{18}{3} $$

$$ t = 6 \mathrm{s} $$

Now we have the time at which the particle's $$x$$-coordinate is $$84 \mathrm{m}$$. Next, we find final velocities in $$x$$ and $$y$$ directions at $$ t = 6 \mathrm{s} $$.

The final velocity in the $$x$$-direction can be found using the formula for velocity with constant acceleration:

$$ v_x = v_{0x} + a_xt $$

$$ v_x = 5 \mathrm{~m/s} + (3 \mathrm{~m/s}^2)(6 \mathrm{s}) $$

$$ v_x = 5 \mathrm{~m/s} + 18 \mathrm{~m/s} $$

$$ v_x = 23 \mathrm{~m/s} $$

Similarly, for the $$y$$-direction:

$$ v_y = v_{0y} + a_yt $$

Since the particle starts from the origin and is only subject to a force after $$t=0$$, its initial velocity in the $$y$$-direction is $$0$$.

$$ v_y = 0 + (2 \mathrm{~m/s}^2)(6 \mathrm{s}) $$

$$ v_y = 12 \mathrm{~m/s} $$

Now we can compute the speed of the particle, which is the magnitude of the velocity vector:

$$ v = \sqrt{v_x^2 + v_y^2} $$

$$ v = \sqrt{(23 \mathrm{~m/s})^2 + (12 \mathrm{~m/s})^2} $$

$$ v = \sqrt{529 + 144} $$

$$ v = \sqrt{673} $$

Therefore, the speed of the particle at the time its $$x$$-coordinate is $$84 \mathrm{m}$$ is $$\sqrt{673} \mathrm{m/s}$$.

So, $$\alpha = 673$$.