Let $$C$$ be a circle with radius $$\sqrt{10}$$ units and centre at the origin. Let the line $$x + y = 2$$ intersects the circle $$C$$ at the points $$P$$ and $$Q$$. Let $$MN$$ be a chord of $$C$$ of length $$2$$ unit and slope $$-1$$. Then, a distance (in units) between the chord $$PQ$$ and the chord $$MN$$ is

Circle $$C: x^2 + y^2 = 10$$. Chord $$PQ$$ on $$x + y = 2$$. Chord $$MN$$ length 2, slope $$-1$$.

Distance of $$PQ$$ from origin.

The line is $$x + y - 2 = 0$$. $$d_1 = \frac{|0+0-2|}{\sqrt{1^2+1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$.

 Distance of $$MN$$ from origin.

$$MN$$ has slope $$-1$$, so it is parallel to $$PQ$$ (equation $$x + y = k$$).

Length of chord $$L = 2\sqrt{r^2 - d_2^2}$$.

$$2 = 2\sqrt{10 - d_2^2} \implies 1 = 10 - d_2^2 \implies d_2 = \sqrt{9} = 3$$.

Distance between chords.

Since they are parallel and on the same side (or opposite), the distance is $$|d_2 - d_1|$$.

Distance = $$\mathbf{3 - \sqrt{2}}$$ (Option A).