We have triangle $∆ABC$

$AB: 2x+3y-1=0 ...\left(1\right)$

Slope $m_1=-\frac{2}{3}$

$BC: x+2y+1=0 ...\left(2\right)$

Slope $m_2=-\frac{1}{2}$

$CA:\hspace{0.17em}ax+by-1=0 ...\left(3\right)$

Slope $m_3=-\frac{a}{b}$

We know, the orthocenter of a triangle is the point of intersection of altitudes passing through the vertices of the triangle.

Here we have altitudes $AD, BE \mathrm{and} CF$

So, the family of line passing through the vertex $B$,

is, $\left(2x+3y-1\right)+\lambda \left(x+2y+1\right)=0 ...\left(4\right)$

As, the orthocentre is origin,

Substituting $\left(x=0 \mathrm{and} y=0\right)$ in $\left(4\right)$, we get $\lambda =1$,

So, the equation of altitude $BE$ is $3x+5y=0 ...\left(5\right)$

Slope $m_{BE}=-\frac{3}{5}$

As, $BE\perp AC$

So, the product of slopes of $BE$ and side $AC=-1$

$\Rightarrow \left(-\frac{a}{b}\right)\times \left(-\frac{3}{5}\right)=-1\Rightarrow 3a=-5b ...\left(6\right)$

And, the family of lines passing through the vertex $A$ is, 

$\left(2x+3y-1\right)+\mu \left(ax+by-1\right)=0 ...\left(7\right)$

As, the orthocentre is origin,

Substituting $\left(x=0 \mathrm{and} y=0\right)$ in $\left(7\right)$, we get, $\mu =-1$

So, the equation of altitude $AD$ is $\left(2-a\right)x+\left(3-b\right)y=0 ...\left(8\right)$

As, $AD$ $\perp$side $BC$.

So, product of their slopes$=-1$

$\left(-\frac{2-a}{3-b}\right)\times \left(-\frac{1}{2}\right)=-1 \Rightarrow \frac{2-a}{2\left(3-b\right)}=-1 \Rightarrow a+2b=8 ...\left(9\right)$

Solving equations $\left(6\right)$ and $\left(9\right)$, we get

$a=-40 \mathrm{and} b=-24$.

Hence, $\frac{1}{a}+\frac{1}{b}=-\frac{1}{40}+\frac{1}{24}=\frac{1}{60}$