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

$$\begin{gathered} A(2,1),B(0,0),C(t,4):t\in [0,4] \\ {{\mathrm{B}}^{\text{'}}}_{ }(0,8)\equiv \text{ image of }\mathrm{B}\text{ w.r.t. }\mathrm{y}=4 \\ \text{ for }\mathrm{AC}+\mathrm{BC}+\mathrm{AB}\text{ to be minimum. } \\ {\mathrm{m}}_{\mathrm{AB}\text{'}}=\frac{-7}{2} \end{gathered}$$

$$\begin{gathered} \text{ line }\mathrm{AB}{\text{'}}_{ }\equiv 7x+2y=16 \\ \mathrm{C}\left(\frac{8}{7},4\right) \\ \mathrm{\beta }=\frac{8}{7} \\ \text{ For max. perimeter ,B=}\left(0,0\right)\text{,C=}\left(4,4\right) \\ \left(\because \text{ Maximization will be possible if }\alpha =0\text{ or }4\right) \\ \mathrm{AB}=\sqrt{5}:\mathrm{BC}=4\sqrt{2},\mathrm{AC}=\sqrt{13} \\ 6\mathrm{\alpha }+21\mathrm{\beta }=24+24=48 \end{gathered}$$