An equation of a plane parallel to the plane $$x - 2y + 2z - 5 = 0$$ and at a unit distance from the origin is

The required plane must be parallel to the given plane $$x-2y+2z-5=0$$.
Two planes are parallel when their normal vectors are proportional. Hence every plane parallel to the given one has the same normal vector $$\mathbf{n}=(1,-2,2)$$ and can be written as

$$x-2y+2z+D=0 \quad -(1)$$
where $$D$$ is a constant still to be determined.

Distance of a plane $$Ax+By+Cz+D=0$$ from the origin $$(0,0,0)$$ is given by the formula

$$\text{Distance}=\frac{|D|}{\sqrt{A^{2}+B^{2}+C^{2}}} \quad -(2)$$

For plane $$-(1)$$, $$A=1,\;B=-2,\;C=2$$, so

$$\sqrt{A^{2}+B^{2}+C^{2}}=\sqrt{1^{2}+(-2)^{2}+2^{2}}=\sqrt{1+4+4}=\sqrt{9}=3.$$ Substitute this into $$-(2)$$ and set the distance equal to the given value $$1$$:

$$\frac{|D|}{3}=1 \;\Longrightarrow\; |D|=3.$$ Hence $$D=3$$ or $$D=-3.$$

Writing the corresponding plane equations:
If $$D=3$$: $$x-2y+2z+3=0$$ (not among the options).
If $$D=-3$$: $$x-2y+2z-3=0$$.

Among the given options, only Option A matches:

Option A which is: $$x - 2y + 2z - 3 = 0$$