For all values of $\theta$, the lines represented by the equation$(2\cos \theta +3\sin \theta )x+(3\cos \theta -5\sin \theta )y-(5\cos \theta -2\sin \theta )=0$ passes through a fixed point whose reflection in the line $x+y=\sqrt{2}$ is

The given equation can be written as $(2x+3y-5)\cos \theta +(3x-5y+2)\sin \theta =0$ $2x+3y-5=0\text{ and }3x-5y+2=0$ for all values of $\theta$. The coordinates of the point p of intersection are (1,1).Let Q(h,k) be the reflection of P(1,1) in the line $x+y=\sqrt{2}......\left(i\right)$ 

Then PQ is perpendicular to (1) and the mid-point of PQ lies on (1).

$\therefore \frac{k-1}{h-1}=1\Rightarrow k=h\text{ and }\frac{h+1}{2}+\frac{k+1}{2}=\sqrt{2}$

$\Rightarrow h=k=\sqrt{2}-1$