Let A (a, b), B(3, 4) and (-6, -8) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P($2a + 3$, $7b + 5$) from the line $2x + 3y – 4 = 0$ measured parallel to the line $x – 2y - 1 = 0$ is

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$\frac{17\sqrt{5}}{7}$

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Let A (a, b), B(3, 4) and (-6, -8) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P( $2a + 3$ , $7b + 5$ ) from the line $2x + 3y – 4 = 0$ measured parallel to the line $x – 2y - 1 = 0$ is

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