Answer = 1/2 × 1 × √3/2 × [ (1^2 + (5√3)^2) / (2 × (2 + 5))^2 ] = √3/4 × 76/196 = √3/4 × 19/49 = 19√3/196

Another Solution

Lemma: If in a $\triangle ABC, D,E$ and $F$ are on $AB, BC, AC$ such that $AD:DB=1:a, BE:EC=1:b$ and $FC:FA=1:c$ Then $[DEF]=\left|\frac{[ABC](1+abc)}{(1+a)(1+b)(1+c)}\right|$ Proof: Let $A(x_1,y_1), B(x_2,y_2), C(x_3,y_3)$ So by section formula we obtain, $D\left(\frac{ax_1+x_2}{a+1},\frac{ay_1+y_2}{a+1}\right), E\left(\frac{bx_2+x_3}{a+1},\frac{by_2+y_3}{b+1}\right), F\left(\frac{cx_3+x_1}{c+1},\frac{cy_3+y_1}{c+1}\right)$ $2[ABC]= \begin{Vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{Vmatrix}$ and $2[DEF]= \begin{Vmatrix} \frac{ax_1+x_2}{a+1} & \frac{ay_1+y_2}{a+1} & 1 \\ \frac{bx_2+x_3}{b+1} & \frac{by_2+y_3}{b+1} & 1 \\ \frac{cx_3+x_3}{a+1} & \frac{cy_3+y_1}{c+1} & 1 \end{Vmatrix}$ \begin{align*} 2[DEF]&=\left|\frac{1}{(1+a)(1+b)(1+c)}\right| \begin{Vmatrix} {ax_1+x_2} & {ay_1+y_2} & {a+1}\\ {bx_2+x_3} & {by_2+y_3} & {b+1} \\ {cx_3+x_3} & {cy_3+y_1} & {c+1} \end{Vmatrix}\\ & =\left|\frac{1}{(1+a)(1+b)(1+c)}\right|\left|\begin{vmatrix} ax_1 & ay_1 & a \\ bx_2 & by_2 & b \\ cx_3 & cy_3 & c \end{vmatrix} + \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right|\\ & =\left|\frac{1}{(1+a)(1+b)(1+c)}\right|\left|abc\begin{vmatrix} {x_1} & {y_1} & {1}\\ {x_2} & {y_2} & {1}\\ {x_3} & {y_3} & {1} \end{vmatrix} + \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right|\\ & =\left|\frac{1}{(1+a)(1+b)(1+c)}\right|\left|abc+1\right| \begin{Vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{Vmatrix}\\ & =\left|\frac{(abc+1)2[ABC]}{(1+a)(1+b)(1+c)}\right| \end{align*} Hence $[DEF]=\left|\frac{[ABC](1+abc)}{(1+a)(1+b)(1+c)}\right|$ Solution: from the figure we obtain $a=5/2, b=5/2, c=5/2$ So $\frac{[PQR]}{[ABC]}=\left|\frac{1+abc}{(1+a)(1+b)(1+c)}\right|=\left|\frac{1+(5/2)(5/2)(5/2)}{(1+5/2)(1+5/2)(1+5/2)}\right|=\frac{19}{49}$ $\implies [PQR]=[ABC]\frac{19}{49}=\left(\frac{\sqrt3 (1)^2}{4}\right)\frac{19}{49}==\boxed{\frac{19\sqrt3}{196}}$