Consider the triangles with vertices A (2, 1), B(0, 0) and C(t, 4),...

Given,

The triangles with vertices $A$\left(2, 1\right), $B\left(0, 0\right)$ and $C\left(t, 4\right), t = \left[0, 4\right]$,

And the maximum and the minimum perimeters of such triangles are obtained at $t = \alpha$ and $t = \beta$ respectively,

Now to minimise $CA + CB$, in below diagram take image of $B$ in $y = 4$,

We get, ${B}^{'} = \left(0, 8\right)$

Now finding, equation of $A{B}^{'}$ we get,

$y - 8 = \dfrac{- 7}{2}\left(x - 0\right)$

Now, putting $y = 4$ in above equation we get,

$- 4 = \dfrac{- 7}{2}\left(x\right)$

$⇒ x = \dfrac{8}{7} ⇒ \beta = \dfrac{8}{7}$

Now, maximum perimeter will be possible if $\alpha = 0$ or $4$

Now taking $\alpha = 0$ we get,

$AB = \sqrt{5}, BC = 4 \text{ and } AC = \sqrt{13}$

Now when $\alpha = 4$ we get,

$AB = \sqrt{5}, BC = 4\sqrt{2} \text{ and } AC = \sqrt{13}$

Now on comparing the perimeter we get, maximum perimeter at $\alpha = 4$

Hence, $6\alpha + 21\beta = 48$