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$Equation of the plane such that foot of altitude drawn from (- 1, 1, 1) to the plane has the$ $coordinate (3, - 2, - 1) is$

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x+y+z=0

4x−3y−2z=20

3x+y−z=8

4x+3y−z=7

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Correct option is B

Clearly, the direction ratios of normal to the plane are 4,−3,−2 . Thus equation of plane will be 4x−3y−2z=d . It will necessarily pass though (3,−2,−1) ie., d=12+6+2=20 Thus the equation of plane is 4x−3y−2z=20

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Equation of the plane such that foot of altitude drawn from (−1,1,1) to the plane has the coordinate (3,−2,−1) is

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$Equation of the plane such that foot of altitude drawn from (- 1, 1, 1) to the plane has the$ $coordinate (3, - 2, - 1) is$