Let P be the image of the point  $Q(7, –2, 5)$ in the line $\mathrm{L}:\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}}{4}$ and $R(5, p, q)$ be a point on L. Then the square of the area of $∆$PQR is_____.

Let $\mathrm{R}(2\mathrm{\lambda }+1,3\mathrm{\lambda }-1,4\mathrm{\lambda })$
$\begin{array}{l}2\mathrm{\lambda }+1=5\\ \mathrm{\lambda }=2\\ \mathrm{R}(5,5,8)\\ \text{ let }\mathrm{T}(2\mathrm{\lambda }+1,3\mathrm{\lambda }-1,4\mathrm{\lambda })\\ \overrightarrow{\mathrm{QT}}=(2\mathrm{\lambda }-6)\widehat{\mathrm{i}}+(3\mathrm{\lambda }+1)\widehat{\mathrm{j}}+(4\mathrm{\lambda }-5)\widehat{\mathrm{k}}\\ \overrightarrow{\mathrm{b}}=2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+4\widehat{\mathrm{k}}\\ \overrightarrow{\mathrm{QT}}\cdot \overrightarrow{\mathrm{b}}=0\\ 4\mathrm{\lambda }-12+9\mathrm{\lambda }+3+16\mathrm{\lambda }-20=0\end{array}$
$\begin{array}{l}\mathrm{\lambda }=1\\ \mathrm{T}(3,2,4)\\ \mathrm{QT}=\sqrt{33}\mathrm{RT}=\sqrt{29}\\ (\text{ area of }\mathrm{\Delta PQR}{)}^2={\left(\frac{1}{2}\sqrt{29}\cdot 2\sqrt{33}\right)}^2\\ =957\end{array}$