JEE MATHEMATICS MCQ: 3D PLANES

1 Easy

1 of 1

Application +4 | -1

Let the equation of the plane, that passes through the point (1, 4, -3) and contains the line of intersection of the planes

3x - 2y + 4z - 7\, = \,0

and

x + 5y - 2z + 9\, = \,0

\alpha x + \beta y + \gamma z + 3\, = \,0,

then

\alpha + \beta + \gamma \,

is equal to

View answer and explanation

Correct option: A

-23

Option A (-23) is correct because it represents the sum of the coefficients

\alpha

\beta

, and

\gamma

in the equation of the plane that passes through the given point and contains the line of intersection of the two planes.

More Questions from 3D PLANES

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\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}

p x-q y+

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b x+a y-z=0

a^2+b^2+c^2+2 a b c

2 x-3 y+9 z=0

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4 x-2 y+7 z+9=0

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\frac{x-4}{1}=\frac{y-7}{5}=\frac{z-4}{4}

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\mathrm{ax}+\mathrm{by} \pm \mathrm{z} \sqrt{\mathrm{a}^2+\mathrm{b}^2} \tan \alpha+\mathrm{c}=0

(3,2,1)

x, y

z

A, B

C

y z

A

z x

B

x y

C