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Let $X={\ell }_B\cap {\ell }_C,$ $Y={\ell }_A\cap {\ell }_C,$ and $Z={\ell }_A\cap {\ell }_B.$ Here is a diagram of the initial position:



Note that triangle $XZY$ is a $45^{\circ }$-$45^{\circ }$-$90^{\circ }$ triangle. Since all three lines rotate at the same rate, the angles between these lines always stay the same, so triangle $XZY$ will always be a $45^{\circ }$-$45^{\circ }$-$90^{\circ }$ triangle.

Let $\alpha =\angle CAZ.$ Depending on the position of the lines, $\angle AZB$ is either $45^{\circ }$ or $135^{\circ }.$ Either way, by the Law of Sines on triangle $ABZ,$ $$\frac{BZ}{\sin \alpha }=\frac{11}{\sin 45^{\circ }},$$so $BZ=11\sqrt{2}\sin \alpha .$



Depending on the positions of the lines, $\angle BCX$ is either $90^{\circ }-\alpha ,$ $\alpha -90^{\circ },$ or $\alpha +90^{\circ }.$ In any case, by the Law of Sines on triangle $BCX,$ $$\frac{BX}{|\sin (90^{\circ }-\alpha )|}=\frac{7}{\sin 45^{\circ }},$$so $BX=7\sqrt{2}|\cos \alpha |.$

Again, depending on the positions of the lines, $XZ$ is the sum or the difference of $BX$ and $BZ,$ which means it is of the form $$\pm 11\sqrt{2}\sin \alpha \pm 7\sqrt{2}\cos \alpha .$$Then $$XY=YZ=\pm 11\sin \alpha \pm 7\cos \alpha .$$By the Cauchy-Schwarz inequality, for any combination of plus signs and minus signs, $$(\pm 11\sin \alpha \pm 7\cos \alpha {)}^2\le (11^2+7^2)({\sin }^2\alpha +{\cos }^2\alpha )=170,$$so $[XYZ]=\frac{X{Y}^2}{2}\le 85.$

We can confirm that equality occurs when $\alpha$ is the obtuse angle such that $\cos \alpha =-\frac{7}{\sqrt{170}}$ and $\sin \alpha =\frac{11}{\sqrt{170}}.$



Therefore, the maximum area of triangle $XYZ$ is $\boxed{85}.$