If the straight lines $2 x+3 y-1=0$, $x+2 y-1=0$ and $a x+b y-1=0$ form a triangle with orthocentre at the origin, then $(a, b)=$

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 $b=8$ $\therefore \quad(a, b)=(-8,8)$.