The point (2, l) is shifted by $3\sqrt{2}$ units parallel $x+y=1$  the line in the direction of increasing ordinate, to reach Q The image of C2 in the line $x+y=1$, is

The equation of a line passing through (2, 1) and parallel to x + y = 1 is

$\frac{x-2}{\cos 3\pi /4}=\frac{y-1}{\sin \pi /4}\text{ or },\frac{x-1}{-1/\sqrt{2}}=\frac{y-1}{1/\sqrt{2}}$

The coordinates of points on this line which are at a distance $3\sqrt{2}$ from (2, 1) are given by

$\frac{x-2}{-1/\sqrt{2}}=\frac{y-1}{1/\sqrt{2}}=\pm 3\sqrt{2}\Rightarrow x=2+3, y=1\pm 3$

But, the ordinate of Q is more than that of P ( 2, 1 ). So, the coordinates of Q are (- 1, 4). The image of Q (- 1, 4) in the line x + y = 1 is given by

$\begin{array}{r}\frac{x+1}{1}=\frac{y-4}{1}=-2\left(\frac{-1+4-1}{1^2+1^2}\right)\\ \frac{x+1}{1}=\frac{y-4}{1}=-2\Rightarrow x=-3,y=2\end{array}$

Thus, the coordinates of the required point are (- 3, 2)