If the two lines $\frac{3}{2}\mathrm{x}+(2\mathrm{a}-1)\mathrm{y}=3$ and 2x+a²y=3 are perpendicular, then the distance of their point of intersection from the point (1, 1) is

The lines $\frac{3}{2}\mathrm{x}+(2\mathrm{a}-1)\mathrm{y}=3,2\mathrm{x}+{\mathrm{a}}^2\mathrm{y}=3$ are perpendicular $\Rightarrow \left(\frac{3}{2}\right)\left(2\right)+(2\mathrm{a}-1){\mathrm{a}}^2=0$
$\Rightarrow 2{\mathrm{a}}^3-{\mathrm{a}}^2+3=0\Rightarrow \mathrm{a}=-1$
Given lines become x- 2y = 2,2x +y = 3. Point of intersection $\mathrm{P}=\left(\frac{8}{5},-\frac{1}{5}\right)$
The distance of P from ( 1 , 1) is  $\sqrt{(8/5-1{)}^2+(-1/5-1{)}^2}=\sqrt{9/25+36/25}=3/\sqrt{5}$