The equation of the containing vectors $\hat{\mathbf{i}}$ and $\hat{\mathbf{i}}+\hat{\mathbf{j}}$
$$
\begin{array}{rlrl}
& & {[r-\hat{\mathbf{i}}, \hat{\mathbf{i}}, \hat{\mathbf{i}}+\hat{\mathbf{j}}]} & =0 \\
\Rightarrow & & (r-\hat{\mathbf{i}}) \cdot[\hat{\mathbf{i}} \times \hat{\mathbf{i}}+\hat{\mathbf{j}}] & =0 \\
\Rightarrow & & (r-\hat{\mathbf{i}}) \hat{\mathbf{k}} & =0 \\
\Rightarrow & & {[(x-1) \hat{\mathbf{i}} \hat{\mathbf{j}}+z \hat{\mathbf{k}}] \hat{\mathbf{k}}=0} \\
\Rightarrow & & & z=0
\end{array}
$$
The equation at the plane containing vectors $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ and $\hat{\mathbf{i}}+\hat{\mathbf{k}}$
$$
\begin{array}{r}
r-(\hat{\mathbf{i}}-\hat{\mathbf{j}}) \cdot[(\hat{\mathbf{i}}-\hat{\mathbf{j}}) \times(\hat{\mathbf{i}}+\hat{\mathbf{k}})]=0 \\
\Rightarrow \quad(r-\hat{\mathbf{i}}+\hat{\mathbf{j}})\langle\hat{\mathbf{i}} \times \hat{\mathbf{i}}+\hat{\mathbf{i}} \times \hat{\mathbf{k}}-\hat{\mathbf{j}} \times \hat{\mathbf{i}}-\hat{\mathbf{j}} \times \hat{\mathbf{k}})=0
\end{array}
$$


$$
\begin{array}{cc}
\Rightarrow & {[(x-1) \hat{\mathbf{i}}+(y+1) \hat{\mathbf{j}}+z \hat{\mathbf{k}}] \cdot(-\hat{\mathbf{j}}+\hat{\mathbf{k}}-\hat{\mathbf{i}})=0} \\
\Rightarrow & -(x-1)-(y+1)+z=0 \\
\Rightarrow & x+y-z=0
\end{array}
$$
Let $\quad a=a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}$
Since, $a$ is parallel to the planes Eqs. (i) and (ii).
$$
\begin{aligned}
& \therefore \quad a_3=0 \text { and } a_1+a_2-a_3=0 \\
& \Rightarrow \quad a_1=-a_2, a_3=0 \quad \therefore \quad a=a_1(\hat{\mathbf{i}}-\hat{\mathbf{j}})
\end{aligned}
$$
There is a vector $b=\hat{\mathbf{i}}-\hat{\mathbf{j}}$ in the direction at $a$ the angle between $a$ or $b$ and $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
$$
\begin{array}{rlrl}
& \therefore & \cos \theta & =\frac{1 \times 1+(-2) \times(-1)}{\sqrt{1+1} \sqrt{1+4+4}} \\
\Rightarrow & & \cos \theta & =\frac{1+2}{\sqrt{2} \times 3}=\frac{3}{3 \sqrt{2}} \Rightarrow \cos \theta=\frac{1}{\sqrt{2}} \\
\Rightarrow & \theta & =\frac{\pi}{4}=45^{\circ}
\end{array}
$$