Answer
$\dfrac{\pi}{4}$
Work Step by Step
The angle between two plane is given by:
$ \theta = \cos ^{-1} (\dfrac{a \cdot b}{|a||b|})$
Here, $a=\lt 1,1,0\gt$ and $b=\lt 2,-1,-2 \gt$
$|a|=\sqrt{1^2+1^2+0^2}=\sqrt{1+1}=\sqrt 2$ and $|b|=\sqrt{2^2+1^2+(-2)^2}=\sqrt 9=3$
Now, $ \theta =\cos ^{-1} (\dfrac{3}{\sqrt 2 (3)})$
or, $=\cos ^{-1} (\dfrac{\sqrt 2}{2})$
or, $=\dfrac{\pi}{4}$
