Equation of line perpendicular to line $a x+b y-1=0$ and passes through the origin is
$$
b x-a y=0
$$
Now, altitude (i) passes through the intersection of lines $2 x+3 y-1=0$, and $x+2 y-1=0$
So,
$$
\begin{aligned}
-b-a & =0 \\
a+b & =0
\end{aligned}
$$
Due to relation (ii), the line $a x+b y-1=0$, becomes
$$
x-y-\frac{1}{a}=0
$$
Now, the altitude perpendicular to the line $2 x+3 y-1=0$, and passes through origin is
$$
3 x=2 y
$$
The point of intersection line (iii) and $x+2 y-1=0$ is $\left(\frac{1}{3}\left(1+\frac{2}{a}\right), \frac{1}{3}\left(1-\frac{1}{a}\right)\right)$ satisfy the line
So, $\quad a=-8$ and $b=8$
$\therefore \quad(a, b)=(-8,8)$.