Exams Joint Entrance Examination Mathematics 3D Geometry Question 237 Question 237 of 521 Medium MCQ
Let the equation of the plane passing through the line of intersection of the planes x + 2 y + a z = 2 x+2y+az=2 x + 2 y + a z = 2 and x − y + z = 3 x-y+z=3 x − y + z = 3 be 5 x − 11 y + b z = 6 a − 1 5x-11y+bz=6a-1 5 x − 11 y + b z = 6 a − 1 . For c ∈ Z c\in\mathbb Z c ∈ Z , if the distance of this plane from the point ( a , − c , c ) (a,-c,c) ( a , − c , c ) is 2 a \dfrac{2}{\sqrt a} a 2 , then a + b c \dfrac{a+b}{c} c a + b is equal to:
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Family: ( x + 2 y + a z − 2 ) + λ ( x − y + z − 3 ) = 0 (x+2y+az-2)+\lambda(x-y+z-3)=0 ( x + 2 y + a z − 2 ) + λ ( x − y + z − 3 ) = 0 . Matching coefficients with 5 x − 11 y + b z = 6 a − 1 5x-11y+bz=6a-1 5 x − 11 y + b z = 6 a − 1 :
\[\frac{1+\lambda}{5}=\frac{2-\lambda}{-11}=\frac{a+\lambda}{b}=\frac{2+3\lambda}{6a-1}.\]
Solving gives λ = − 7 2 , a = 3 , b = 1 \lambda=-\dfrac72,\ a=3,\ b=1 λ = − 2 7 , a = 3 , b = 1 .
Distance condition:
\[\frac{2}{\sqrt a}=\frac{|5a+11c+bc-6a+1|}{\sqrt{25+121+1}} \implies c=-1.\]
\[\frac{a+b}{c}=\frac{3+1}{-1}=-4.\] Answer the question to unlock