Consider the triangles with vertices $$A(2, 1)$$, $$B(0, 0)$$ and $$C(t, 4)$$, $$t = [0, 4]$$. If the maximum and the minimum perimeters of such triangles are obtained at $$t = \alpha$$ and $$t = \beta$$ respectively, then $$6\alpha + 21\beta$$ is equal to _____.

$$AB = \sqrt{(2 - 0)^2 + (1 - 0)^2} = \sqrt{4 + 1} = \sqrt{5}$$

$$BC = \sqrt{(t - 0)^2 + (4 - 0)^2} = \sqrt{t^2 + 16}$$

$$CA = \sqrt{(t - 2)^2 + (4 - 1)^2} = \sqrt{(t - 2)^2 + 9}$$

Perimeter, $$P(t) = \sqrt{5} + \sqrt{t^2 + 16} + \sqrt{(t - 2)^2 + 9}$$

The sum of distances $$BC + CA$$ represents the path length from $$B(0,0)$$ to the line $$y=4$$ at $$C(t,4)$$, and then to $$A(2,1)$$.

Using the reflection principle to find the shortest path, we reflect point $$A(2,1)$$ across the line $$y = 4$$. Let this image point be $$A'$$.

The $$y$$-coordinate of $$A'$$ is $$4 + (4 - 1) = 7$$, so $$A' = (2,7)$$.

The shortest path occurs when $$B(0,0)$$, $$C(t,4)$$, and $$A'(2,7)$$ are collinear.

$$\text{Slope of } BC = \text{Slope of } BA'$$

$$\frac{4 - 0}{t - 0} = \frac{7 - 0}{2 - 0} \implies \frac{4}{t} = \frac{7}{2}$$

$$7t = 8 \implies t = \frac{8}{7}$$

Since $$\frac{8}{7} \in [0,4]$$, the minimum perimeter is achieved exactly at this point: $$\beta = \frac{8}{7}$$

Since $$P(t)$$ is a convex function, its maximum value over a closed interval $$[0,4]$$ must occur at one of the boundary endpoints, either $$t = 0$$ or $$t = 4$$.

At $$t = 0: $$

$$BC = \sqrt{0^2 + 16} = 4$$

$$CA = \sqrt{(0 - 2)^2 + 9} = \sqrt{4 + 9} = \sqrt{13} \approx 3.61$$

$$BC + CA = 4 + 3.61 = 7.61$$

At $$t = 4: $$

$$BC = \sqrt{4^2 + 16} = \sqrt{32} = 4\sqrt{2} \approx 5.66$$

$$CA = \sqrt{(4 - 2)^2 + 9} = \sqrt{4 + 9} = \sqrt{13} \approx 3.61$$

$$BC + CA = 5.66 + 3.61 = 9.27$$

Hence, $$\alpha = 4$$

$$\text{Value} = 6(4) + 21\left(\frac{8}{7}\right)$$

$$\text{Value} = 24 + 3 \times 8 = 24 + 24 = 48$$