The point represented by $$2 + i$$ in the Argand plane moves 1 unit eastwards, then 2 units northwards and finally from there $$2\sqrt{2}$$ units in the south-west wards direction. Then its new position in the Argand plane is at the point represented by:

We begin with the complex number $$z_0 = 2 + i$$. In the Argand (complex) plane, the real part $$2$$ gives the horizontal (east-west) coordinate and the imaginary part $$1$$ gives the vertical (north-south) coordinate.

First, the point moves $$1$$ unit eastwards. Eastwards means we add $$1$$ to the real part while the imaginary part stays the same. Therefore

$$ z_1 \;=\; (2 + 1) + i \;=\; 3 + i. $$

Next, the point moves $$2$$ units northwards. Northwards means we add $$2$$ to the imaginary part while the real part remains unchanged. So

$$ z_2 \;=\; 3 + (1 + 2)i \;=\; 3 + 3i. $$

Finally, the point moves $$2\sqrt{2}$$ units in the south-west direction. “South-west” indicates a direction exactly halfway between south and west, that is, at an angle of $$225^\circ$$ from the positive real axis (or $$45^\circ$$ below the negative real axis). In such a direction the horizontal and vertical components are equal in magnitude but both negative.

Using the trigonometric resolution of a vector, we write the components:

$$ \text{Horizontal component} \;=\; 2\sqrt{2}\,\cos 225^\circ, \qquad \text{Vertical component} \;=\; 2\sqrt{2}\,\sin 225^\circ. $$

We know the exact values $$\cos 225^\circ = -\dfrac{\sqrt{2}}{2}$$ and $$\sin 225^\circ = -\dfrac{\sqrt{2}}{2}$$, hence

$$ 2\sqrt{2}\,\cos 225^\circ =\; 2\sqrt{2}\left(-\dfrac{\sqrt{2}}{2}\right) =\; -2,\\[4pt] 2\sqrt{2}\,\sin 225^\circ =\; 2\sqrt{2}\left(-\dfrac{\sqrt{2}}{2}\right) =\; -2. $$

Thus the south-west shift subtracts $$2$$ from the real part and $$2$$ from the imaginary part. Applying this to $$z_2$$ gives

$$ z_3 =\;(3 - 2) + (3 - 2)i =\; 1 + i. $$

This complex number $$1 + i$$ represents the new position in the Argand plane.

Hence, the correct answer is Option A.