$\begin{array}{1 1} (a)\;x-2y+2z-3=0\:\:\:\qquad\:\:(b)\:\:x-2y+2z+1=0\:\:\qquad\:\:(c)\:\:x-2y+2z-1=0\:\:\qquad\:\:(d)\:\:x-2y+2z+5=0 \end{array} $

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- Equation of any plane parallel to the plane $ax+by+cz+d=0$ is $ax+by+cz+d_1=0$
- Distance of the plane $ax+by+cz+d=0\:\:$from origin is $ \bigg|\large\frac{d}{\sqrt {a^2+b^2+c^2}}\bigg|$

Let the eqn. of the plane parallel to $x-2y+2z-5=0$ be $x-2y+2z+d=0$

Given that this plane is at unit distance from origin.

$\Rightarrow\:\bigg|\large\frac{d}{\sqrt {1+4+4}}\bigg|=1$

$\Rightarrow\:d=\pm3$

$\therefore $ Required eqn. of the plane is $x-2y+2z-3=0$

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