A particle starts from origin at $$t = 0$$ with a velocity $$5\hat{i} \text{ m s}^{-1}$$ and moves in $$x - y$$ plane under action of a force which produces a constant acceleration of $$(3\hat{i} + 2\hat{j}) \text{ m s}^{-2}$$. If the $$x$$-coordinate of the particle at that instant is $$84$$ m, then the speed of the particle at this time is $$\sqrt{\alpha} \text{ m s}^{-1}$$. The value of $$\alpha$$ is _______.

Initial velocity: $$u_x = 5$$ m/s, $$u_y = 0$$.

Acceleration: $$a_x = 3$$ m/s$$^2$$, $$a_y = 2$$ m/s$$^2$$.

The x-coordinate: $$x = u_x t + \frac{1}{2}a_x t^2 = 5t + \frac{3}{2}t^2$$

Setting $$x = 84$$:

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

$$3t^2 + 10t - 168 = 0$$

Using the quadratic formula:

$$t = \frac{-10 + \sqrt{100 + 2016}}{6} = \frac{-10 + \sqrt{2116}}{6} = \frac{-10 + 46}{6} = \frac{36}{6} = 6$$ s

Velocity components at $$t = 6$$ s:

$$v_x = u_x + a_x t = 5 + 3(6) = 23$$ m/s

$$v_y = u_y + a_y t = 0 + 2(6) = 12$$ m/s

Speed: $$v = \sqrt{v_x^2 + v_y^2} = \sqrt{529 + 144} = \sqrt{673}$$ m/s

So $$\alpha = 673$$.

The answer is $$\boxed{673}$$.