The problem is to find the equation of a plane (let's call it $A$) that is perpendicular to the $yz$ plane, containing the point $P(2,1,1)$, and making an angle of $\cos^{-1} \frac{2}{3}$ with the plane $2x - y + 2z - 3 = 0$ (let's call it $B$).
First, I know that the equation perpendicular to the $yz$ plane will have $x=0$. Second, I think that the angle between planes $A$ and $B$ will still be $\cos^{-1} \frac{2}{3}$ where $x=0$, ie within the $yz$ plane. So, the equation of the line intersecting planes $B$ and $yz$ is
$$ -y + 2z - 3 = 0 $$
And let the line intersecting planes $A$ and $yz$ be
$$ b(y-1) + c(z-1) = 0 $$
Third, I also think that the angle between the normal lines perpendicular to those equations should still be $\cos^{-1} \frac{2}{3}$. So, I use this formula and make a few calculations
$$ \vec{B}\cdot \vec{A} = \|\vec{B}\|\|\vec{A}\|\cos \theta \\ \langle 0,-1,2\rangle \cdot \langle 0,b,c\rangle = \|\langle 0,-1,2 \rangle \|\|\langle 0,b,c \rangle \| \frac{2}{3} \\ 0\cdot 0 + (-1)\cdot b + 2\cdot c = \sqrt{0^2+(-1)^2+2^2}\sqrt{0^2+b^2+c^2}\frac{2}{3} \\ -b + 2c = 2 \sqrt{b^2+c^2} \\ \frac{b^2}{4} - bc + c^2 = b^2 + c^2 \\ 4c = -3b \\ $$
This looks close to a solution but I can't seem to reach the actual solutions, because there are actually two, of $4y-3z-1=0$ and $z=1$.
With the relationship you found between $b$ and $c$, as well as $x=0$, you have
$$ ax + by + cz + d = 0 \\ a(0) + by + (\frac{-3b}{4})z + d = 0 \\ 4by - 3bz + 4d = 0 \\ $$
With the components of $(2,1,1)$, you then have
$$ 4b(1) - 3b(1) + 4d = 0 \\ b = -4d \\ $$
With this relationship between $b$ and $d$, you finally have
$$ 4by - 3bz + 4(\frac{-b}{4}) = 0 \\ 4y - 3z -1 = 0 \\ $$
Note that you made a mistake when calculating $\| \vec{B} \|$ as $\sqrt{0^2+(-1)^2+2^2} = \sqrt{5}$. Instead, you should have $\sqrt{2^2+(-1)^2+2^2} = \sqrt{9} = 3$ which cancels nicely with the denominator. You did this anyway which is a case where two wrongs make a right.