Shortes distance of a point $\left( {{x_1},{y_1}} \right)$ from line $ax + by = c$ is

$d = \frac{{a{x_1} + b{y_1} – c}}{{\sqrt {{a^2} + {b^2}} }}$

Now shortest distance of $P(1,2)$ from $3x+4y=9$ is

$PC = d = \frac{{3\left( 1 \right) + 4\left( 2 \right) – 9}}{{\sqrt {{3^2} + {4^2}} }} = \frac{2}{5}$

Given that $\Delta APB$ is equilateral triangle

Let $'a'$ be its side

then $PB = a,Cb = \frac{a}{2}$

Now,In $\Delta PCB,{\left( {PB} \right)^2} = {\left( {PC} \right)^2} + {\left( {CB} \right)^2}$

(By Pythagoras theorem )

${a^2} = {\left( {\frac{2}{5}} \right)^2} + \frac{{{a^2}}}{4}$

${a^2} – \frac{{{a^2}}}{4} = \frac{4}{{25}} \Rightarrow \frac{{3{a^2}}}{4} = \frac{4}{{25}}$

${a^2} = \frac{{16}}{{75}} \Rightarrow a = \sqrt {\frac{{16}}{{75}}}  = \frac{4}{{5\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} = \frac{{4\sqrt 3 }}{{15}}$

$\therefore $ Lengh of Equilateral triangle $(a) = \frac{{4\sqrt 3 }}{5}$