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If the orthocentre of the triangle formed by the lines $2 x+3 y-1=0, x+2 y-1=0$ and $a x+b y-1=0$, is the centroid of another triangle, whose circumcentre and orthocentre respectively are $(3,4)$ and $(-6,-8)$, then the value of $|a-b|$ is_______
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16
$\begin{aligned} & 2 x+3 y-1=0 \\ & x+2 y-1=0 \\ & a x+b y-1=0\end{aligned}$
$\begin{aligned} & \left(\frac{6-6}{3}, \frac{8-8}{3}\right) \\ & =(0,0)\end{aligned}$
$\begin{aligned} & a x+b y-1=0 \\ & \left(\frac{1-0}{-1-0}\right)\left(\frac{-a}{b}\right)=-1 \\ & \Rightarrow-\mathrm{a}=\mathrm{b} \\ & \Rightarrow \quad \mathrm{ax}-\mathrm{ay}-1=0 \\ & a x-a\left(1-\frac{2 x}{3}\right)-1 \\ & x\left(a+\frac{2 a}{3}\right)=\frac{a}{3} \\ & \mathrm{x}=\frac{\mathrm{a}+3}{5 \mathrm{a}} \\ & 2\left(\frac{a+3}{5 a}\right)+3 y-1=0 \\ & y=\frac{1-\frac{2 a+6}{5 a}}{3}=\frac{3 a-6}{3 \times 5 a} \\ & y=\frac{a-2}{5 a} \\ & \frac{\left(\frac{a-2}{5 a}\right)}{\left(\frac{a+3}{5 a}\right)}=2 \Rightarrow a-2=2 a+6 \\ & a=-8 \\ & \mathrm{~b}=8 \\ & -8 \mathrm{x}+8 \mathrm{y}-1=0 \\ & |a-b|=16 \\ & \end{aligned}$
$\begin{aligned} & \left(\frac{6-6}{3}, \frac{8-8}{3}\right) \\ & =(0,0)\end{aligned}$
$\begin{aligned} & a x+b y-1=0 \\ & \left(\frac{1-0}{-1-0}\right)\left(\frac{-a}{b}\right)=-1 \\ & \Rightarrow-\mathrm{a}=\mathrm{b} \\ & \Rightarrow \quad \mathrm{ax}-\mathrm{ay}-1=0 \\ & a x-a\left(1-\frac{2 x}{3}\right)-1 \\ & x\left(a+\frac{2 a}{3}\right)=\frac{a}{3} \\ & \mathrm{x}=\frac{\mathrm{a}+3}{5 \mathrm{a}} \\ & 2\left(\frac{a+3}{5 a}\right)+3 y-1=0 \\ & y=\frac{1-\frac{2 a+6}{5 a}}{3}=\frac{3 a-6}{3 \times 5 a} \\ & y=\frac{a-2}{5 a} \\ & \frac{\left(\frac{a-2}{5 a}\right)}{\left(\frac{a+3}{5 a}\right)}=2 \Rightarrow a-2=2 a+6 \\ & a=-8 \\ & \mathrm{~b}=8 \\ & -8 \mathrm{x}+8 \mathrm{y}-1=0 \\ & |a-b|=16 \\ & \end{aligned}$
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