Let $P_1$ denotes the equation of a plane to which the vector $\overline{i}+\overline{j}$ is normal and which contains the line L whose equation is $\overline{r}=\overline{i}+\overline{j}+\overline{k}+\lambda (\overline{i}-\overline{j}-\overline{k})$ and $P_2$ denotes the equation of plane containing the line L and a point whose position vector $\overline{j}.$ which of the following holds good?

Plane $P_1$ contain the line $\overline{r}=\overline{i}+\overline{j}+\overline{k}+\lambda (\overline{i}-\overline{j}-\overline{k})$  
It contains the point $\overline{i}+\overline{j}+\overline{k}$ and is normal to vector $\overline{i}+\overline{j}$ 
Hence equation of plane $P_1$ is $(\overline{r}-(\overline{i}+\overline{j}+\overline{k})).(\overline{i}+\overline{j})=0\Rightarrow x+y=2$  
Plane $P_2$ contains the line $\overline{r}=\overline{i}+\overline{j}+\overline{k}+\lambda (\overline{i}-\overline{j}-\overline{k})$ and point $\overline{j}$ 
Equation of Plane $P_2$ is $\left|\begin{array}{ccc}x-0& y-1& z-0\\ 1-0& 1-1& 1-0\\ 1& -1& -1\end{array}\right|$ =0 $\to x+2y-z=2$ 
$\theta$ is angle between $P_1$ and $P_2$ then $\cos \theta =\frac{n_1.n_2}{\left|n_1\right|\left|n_2\right|}=\frac{\sqrt{3}}{2}\to \theta =\frac{\pi }{6}={\cot }^{-1}\sqrt{3}$