A plane passes through a point (2,3,- 4), and it is perpendicular to the line with direction ratios (2,3,-1). The equation of the plane is .

2x + 3y - z = 13

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2x + 3y - z = 17

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3x + 2y - z = 13

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3x + 2y - z = 17

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Solution

The correct option is B 2x + 3y - z = 17
We know the coordinates of a point and also the direction cosines of the perpendicular (normal). Equation of plane in point normal form is

$a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$

Where a, b, c are the direction ratios of the normal and $(x_{1}, y_{1}, z_{1}$ ) are the coordinates of the point given.

So the equation will be $2 (x - 2) + 3 (y - 3) - 1 (z + 4) = 0$

Or 2x + 3y - z = 17

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