Let $\quad l_1 \equiv 2 x+3 y=5$
Since, line $A B \perp I_1$
$\therefore$ Slope of $l_1$ is
$$
\begin{array}{r}
m_1 \text { say }=\frac{-2}{3} \\
\therefore \text { Slope of } A B=\frac{-1}{(-2 / 3)}=\frac{3}{2}
\end{array}
$$


Equation of line $A B$ is
$$
\begin{array}{cc}
& \left(y-\frac{1}{3}\right)=\frac{3}{2}(x-1) \\
\Rightarrow \quad & 3 x-2 y=\frac{7}{3}
\end{array}
$$
Equation of line $I_1$ is
$$
2 x+3 y=5
$$
From Eqs. (i) and (ii), we get
$$
\Rightarrow \quad \begin{array}{rl}
13 x & =17 \\
x & x=\frac{17}{13}
\end{array}
$$
From Eq. (i), we get
$$
3 y=5-\frac{34}{13} \Rightarrow y=\frac{65-34}{13 \times 3}=\frac{31}{39}
$$
So, mid-point $P \rightarrow\left(\frac{17}{13}, \frac{31}{39}\right)$
Coordinate of point $B$
$$
\begin{aligned}
& =\left(\frac{17}{13} \times 2-1, \frac{31}{39} \times 2-\frac{1}{3}\right) \\
& =\left(\frac{34-13}{13}, \frac{62-13}{39}\right)=\left(\frac{21}{13}, \frac{49}{39}\right)
\end{aligned}
$$