$$\begin{gathered} \text{ Let two vertices of a triangle }\mathrm{ABC}\text{ be }(2,4,6)\text{ and }(0,-2,-5)\text{, and its centroid be }(2,1,-1)\text{. If the image of the} \\ \text{third vertex in the plane }\mathrm{x}+2\mathrm{y}+4\mathrm{z}=11\text{ is }(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma })\text{, then }\mathrm{\alpha \beta }+\mathrm{\beta \gamma }+\mathrm{\gamma \alpha }\text{ is equal to } \end{gathered}$$

$$\begin{gathered} \mathrm{A}(2,4,6),\mathrm{B}(0,-2,-5)\mathrm{G}(2,1,-1) \\ \text{ Let the third vertex }\mathrm{C}(\mathrm{x},\mathrm{y},\mathrm{z}) \end{gathered}$$

$$\begin{gathered} \text{ Third vertex }=(4,1,-4) \\ \mathrm{Let} (\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }) \mathrm{be} \mathrm{the} \text{ image of vertex C in the plane x+2y+4z=11} \\ \frac{\alpha -4}{1}=\frac{\beta -1}{2}=\frac{\gamma +4}{4}=\frac{-2(4+2-16-11)}{21} \\ \therefore \text{ image }=(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma })=(6,5,4) \\ \mathrm{\alpha \beta }+\mathrm{\beta \gamma }+\mathrm{\gamma \alpha }=74 \end{gathered}$$