Concept:A plane through the intersection of two given planes belongs to their family, and parallel to a line means their normal must be perpendicular to the line's direction vector.
Explanation:The family of planes through the intersection of
2x−y+z−3=0 and
4x−3y+5z+9=0 is:
(2x−y+z−3)+λ(4x−3y+5z+9)=0So, the normal vector of the required plane is:
(2+4λ,−1−3λ,1+5λ)The given line has direction vector
(2,4,5).
Since the plane is parallel to the line, the normal vector is perpendicular to the line's direction vector:
(2+4λ)(2)+(−1−3λ)(4)+(1+5λ)(5)=0Simplifying:
4+8λ−4−12λ+5+25λ=05+21λ=0λ=−215​Substitute
λ=−215​ into the family of planes:
(2x−y+z−3)−215​(4x−3y+5z+9)=0Multiplying by
21 and simplifying:
42x−21y+21z−63−20x+15y−25z−45=022x−6y−4z−108=0Dividing by
2:
11x−3y−2z−54=0Answer:The required plane is
11x−3y−2z−54=0, which matches Option A.