We use Co-ordinate Geometry to solve this Problem.

Without ant loss of generality, we may assume that the centre of the

circle is the origin.

A point #P, pi/4# radian on a circle of radius #r# have co-ordinates

#P(rcos(pi/4), rsin(pi/4))=P(r/sqrt2,r/sqrt2)#.

Similarly, a point #Q, pi/3# radians on the circle is #Q(r/2,(rsqrt3)/2)#.

Given that #PQ=3 rArr PQ^2=9#

#rArr (r/sqrt2-r/2)^2+(r/sqrt2-(rsqrt3)/2)^2=9#.

#rArr ((rsqrt2)/2-r/2)^2+((rsqrt2)/2-(rsqrt3)/2)^2=9#.

#rArr{r/2(sqrt2-1)}^2+{r/2(sqrt2-sqrt3)}^2=9#.

#rArr r^2/4{(2-2sqrt2+1)+(2-2sqrt6+3)}=9#

#rArr r^2/4(8-2sqrt2-2sqrt6)=9#.

#rArr r^2=36/(8-2sqrt2-2sqrt6)#

Taking, #sqrt2~=1.414, sqrt6~=2.449#, we have,

#r^2=36/(8-2.828-4.898)=36/0.274=131.39#, so,

the area of the circle# = pir^2=3.14*131.39=412.56#