Let the co-ordinates of one vertex of $$\triangle ABC$$ be $$A(0, 2, \alpha)$$ and the other two vertices lie on the line $$\frac{x+\alpha}{5} = \frac{y-1}{2} = \frac{z+4}{3}$$. For $$\alpha \in \mathbb{Z}$$, if the area of $$\triangle ABC$$ is $$21$$ sq. units and the line segment $$BC$$ has length $$2\sqrt{21}$$ units, then $$\alpha^2$$ is equal to ______.

Given that the area is 21 sq. units and the base (length of segment $$BC$$) is $$2\sqrt{21}$$ units, we can find the height $$h$$, which represents the perpendicular distance from vertex $$A(0, 2, \alpha)$$ to the line containing $$B$$ and $$C$$:

$$21 = \frac{1}{2} \times 2\sqrt{21} \times h \implies h = \sqrt{21}$$

The equation of the line is:

$$\frac{x+\alpha}{5} = \frac{y-1}{2} = \frac{z+4}{3}$$

This line passes through the point $$P(-\alpha, 1, -4)$$ and has the direction vector:

$$\vec{d} = 5\hat{i} + 2\hat{j} + 3\hat{k}$$

The vector joining $$P$$ to $$A(0, 2, \alpha)$$ is:

$$\vec{PA} = (0 - (-\alpha))\hat{i} + (2 - 1)\hat{j} + (\alpha - (-4))\hat{k} = \alpha\hat{i} + \hat{j} + (\alpha + 4)\hat{k}$$

The perpendicular distance $$h$$ from point $$A$$ to the line is given by:

$$h = \frac{|\vec{PA} \times \vec{d}|}{|\vec{d}|}$$

First, we compute the cross product $$\vec{PA} \times \vec{d}$$:

$$\vec{PA} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 1 & \alpha+4 \\ 5 & 2 & 3 \end{vmatrix}$$

$$\vec{PA} \times \vec{d} = \hat{i}(3 - 2(\alpha+4)) - \hat{j}(3\alpha - 5(\alpha+4)) + \hat{k}(2\alpha - 5)$$

$$\vec{PA} \times \vec{d} = (-2\alpha - 5)\hat{i} + (2\alpha + 20)\hat{j} + (2\alpha - 5)\hat{k}$$

Now, we square the magnitude of the cross product:

$$|\vec{PA} \times \vec{d}|^2 = (-2\alpha - 5)^2 + (2\alpha + 20)^2 + (2\alpha - 5)^2$$

$$|\vec{PA} \times \vec{d}|^2 = (4\alpha^2 + 20\alpha + 25) + (4\alpha^2 + 80\alpha + 400) + (4\alpha^2 - 20\alpha + 25)$$

$$|\vec{PA} \times \vec{d}|^2 = 12\alpha^2 + 80\alpha + 450$$

The square of the magnitude of the direction vector $$\vec{d}$$ is:

$$|\vec{d}|^2 = 5^2 + 2^2 + 3^2 = 25 + 4 + 9 = 38$$

Using the relation $$h^2 = 21$$:

$$21 = \frac{12\alpha^2 + 80\alpha + 450}{38}$$

$$798 = 12\alpha^2 + 80\alpha + 450$$

$$12\alpha^2 + 80\alpha - 348 = 0$$

Dividing the entire equation by 4 to simplify:

$$3\alpha^2 + 20\alpha - 87 = 0$$

Solving this quadratic equation for $$\alpha$$ using the quadratic formula:

$$\alpha = \frac{-20 \pm \sqrt{20^2 - 4(3)(-87)}}{2(3)}$$

$$\alpha = \frac{-20 \pm \sqrt{400 + 1044}}{6}$$

$$\alpha = \frac{-20 \pm \sqrt{1444}}{6} = \frac{-20 \pm 38}{6}$$

This yields two potential roots for $$\alpha$$:

$$\alpha = \frac{18}{6} = 3 \quad \text{or} \quad \alpha = \frac{-58}{6} = -\frac{29}{3}$$

Since we are given that $$\alpha \in \mathbb{Z}$$ (an integer), we get $$\alpha = 3$$.

Conclusion:

The value of $$\alpha^2$$ is equal to 9.