If the straight line 2 x+3 y-1=0, x+2 y-1=0 and a x+b y-1=0 form a...

Here, point $A$ is the intersection of line $A B$ and $A C$ so equation of line passing through $A$. $(x+2 y-1)+\lambda(2 x+3 y-1)=0$.....(i) This line passes through the orthocentre $(0,0)$, then $-1-\lambda=0\Rightarrow \lambda=-1$ On substituting $\lambda=-1$ in Eq. $(i)$, we get $x+y= 0 as the equation of$A D$. Since$A D \perp B C$, therefore <img src="/api/proxy-image?url=https$-1 \times-$\frac{a}{b}=$-1\Rightarrow \quad a+b=0$...(ii) Similarly, by applying the condition that$B E$is perpendicular to$C A$, we get$a+2 b=8$....(iii) Now, solving Eqs. (ii) and (iii), we get$a=-8$,$b=8