The equation to the plane bisecting the line segment joining $(- 3, 3, 2), (9, 5, 4)$ and perpendicular to the line segment is

x - y + 4_{Z} - 13 = 0

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2 x - 2 y + 7 z - 23 = 0

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x - 7 y + 2_{Z} - 1 = 0

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6 x + y + z - 25 = 0

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Solution

The correct option is B $6 x + y + z - 25 = 0$
The point of bisection is $(3, 4, 3)$ .
The d.r's of the line joining the $2$ points are $(9 + 3, 5 - 3, 4 - 2) = (12, 2, 2)$
These are also the d.r's of the normal to the plane.
Hence, the equation of the plane is $12 x + 2 y + 2 z = d$
Since, it passes through $(3, 4, 3)$ .
$∴ 12 (3) + 2 (4) + 2 (3) = d$
$∴ d = 50$
Hence, equation of the plane is
$12 x + 2 y + 2 z = 50$
$6 x + y + z = 25$

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The equation to the plane bisecting the line segment joining (−3,3,2),(9,5,4) and perpendicular to the line segment is

The equation to the plane bisecting the line segment joining $(- 3, 3, 2), (9, 5, 4)$ and perpendicular to the line segment is