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(C) The lines $x + 2y = 3$ and $3x - 2y = 5$ are the perpendicular bisectors of sides $AB$ and $BC$ respectively.Solving these two lines gives the cir

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$5$

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(C) The lines $x + 2y = 3$ and $3x - 2y = 5$ are the perpendicular bisectors of sides $AB$ and $BC$ respectively.Solving these two lines gives the circumcentre $H$ of $	riangle ABC$:$x + 2y = 3$$3x - 2y = 5$Adding them gives $4x = 8 \Rightarrow x = 2$. Substituting $x=2$ gives $2 + 2y = 3$ $\Rightarrow 2y = 1$ $\Rightarrow y = \frac{1}{2}$.So,the circumcentre $H$ is $\left(2, \frac{1}{2}ight)$.The origin $O(0, 0)$ is given as the orthocentre.The point $P(a, b)$ inside the triangle such that $	ext{Area}(	riangle PAB) = 	ext{Area}(	riangle PBC) = 	ext{Area}(	riangle PCA)$ is the centroid of the triangle.In any triangle,the centroid $G$ divides the line segment joining the orthocentre $O$ and the circumcentre $H$ in the ratio $2:1$.Using the section formula for $P(a, b)$ dividing $OH$ in ratio $2:1$:$P = \left( \frac{2(2) + 1(0)}{2+1}, \frac{2(1/2) + 1(0)}{2+1} ight) = \left( \frac{4}{3}, \frac{1}{3} ight)$.Thus,$a = \frac{4}{3}$ and $b = \frac{1}{3}$.$a + b = \frac{4}{3} + \frac{1}{3} = \frac{5}{3}$.$3(a + b) = 3 	imes \frac{5}{3} = 5$.

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