$$
\text { Given points } A(1,0,0) \text { and } B(0,0,1)
$$


Direction ratios of the line joining points are
$$
\left(x_2-x_1\right),\left(y_2-y_1\right),\left(z_2-z_1\right)
$$
i.e. $(-1,1,1)$. So, $a=-1, b=1$ and $c=1$
$\therefore$ Equation of plane
$$
\begin{aligned}
& (x-1) a+(y-0) b & +(z-0) c=0 \\
\Rightarrow & -(x-a)+z & =0 \\
\Rightarrow & \pi & =-x+0 y+z+a=0 \\
\Rightarrow & x+y+z & =6
\end{aligned}
$$
[angle between these two planes]
$$
\begin{array}{ll}
\Rightarrow \quad & \cos \theta=\frac{(-1)(1)+0+(1)(1)}{\sqrt{1+0+1} \sqrt{1+1+1}} \\
\Rightarrow & {\left[\begin{array}{c}
\because \cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}
\end{array}\right]} \\
& \cos \theta=\frac{-1+1}{\sqrt{2} \cdot \sqrt{3}} \Rightarrow 0 \Rightarrow \theta=\frac{\pi}{2}
\end{array}
$$