Let Q and R be two points on the line $\frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}-1}{2}$ at a distance $\sqrt{26}$ from the point P(4,2,7). Then the square of the area of the triangle PQR is ________

$$\begin{gathered} \text{ Let }\left(2\lambda -1,3\lambda -2,2\lambda +1\right)\text{ be any point on the line } \\ and P\left(4,2,7\right) \\ \begin{array}{r}\Rightarrow (2\lambda -5{)}^2+(3\lambda -4{)}^2+(2\lambda -6{)}^2=26\\ \therefore \lambda =1,3\\ \Rightarrow \mathrm{Q}(1,1,3);\mathrm{R}(5,7,7);\mathrm{P}(4,2,7)\\ \boxed{\mathrm{Area} \mathrm{of} ∆\mathrm{PQR}=1/2|\overrightarrow{\mathrm{PQ}}\times \overrightarrow{\mathrm{PR}}|}\\ \overrightarrow{\mathrm{PQ}}\times \overrightarrow{\mathrm{PR}}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -3& -1& -4\\ 1& 5& 0\end{array}\right|=\left(20,-4,-14\right)\\ \mathrm{Area}=\frac{1}{2}|\overrightarrow{\mathrm{PQ}}\times \overrightarrow{\mathrm{PR}}|=\frac{1}{2}\sqrt{400+16+196}\\ =\frac{{\displaystyle \sqrt{612}}}{2}=\frac{2\sqrt{153}}{2}==\sqrt{153}\end{array} \end{gathered}$$