The base of an equilateral triangle is along the line given by 3...

Shortest distance of a point $\left(x_1, y_1\right)$ from line $a x+b y=c$ is $d=\left|\frac{a x_1+b y_1-c}{\sqrt{a^2+b^2}}\right|$ Now shortest distance of $\mathrm{P}(1,2)$ from $3 x+4 y=9$ is $\mathrm{PC}=$d=\left|\frac{3(1)+4(2)-9}{\sqrt{3^2+4^2}}\right|=\frac{2}{5}$Given that$\triangle \mathrm{APB}$is an equilateral triangle Let '$a$' be its side then$\mathrm{PB}=a, \mathrm{CB}=\frac{a}{2}$Now, In$\Delta \mathrm{PCB},(\mathrm{PB})^2=(\mathrm{PC})^2+(\mathrm{CB})^2$(By Pythagoras theorem) a²=$\left(\frac{2}{5}\right)^$2+$\frac{a^2}{4}a^2-\frac{a^4}{4}=\frac{4}{25} \Rightarrow \frac{3 a^2}{4}=\frac{4}{25}a^2=\frac{16}{75} \Rightarrow a=\sqrt{\frac{16}{75}}=\frac{4}{5 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{4 \sqrt{3}}{15}