If the point `(2 , alpha , beta) ` lies on the plane which passes through the points `(3,4,2) and (7,0,6)` and is perpendicular to the plane `2x-5y=15,` then `2 alpha - 3 beta` is equal to:

Options

A. 17
B. 7
C. 12
D. 5

Answer

B

Explanation

Normal vector of plane passing through `(2, alpha , beta ), (2,4,2) and (7,0,6)` is
`vecn = |{:(hati, hatj, hatk),(-5, alpha, beta -6),(4,-4, 4):}|therefore " " vecn = hati ( 4 alpha - 4 beta -24) + hatj(4 beta-24)+ hatk (20-4 alpha )`
`therefore` This plane is perpendicular `o 2x -5y =15therefore` Dot product of their normal is 0.
`implies 2 (4 alpha + 4 beta -24) -5(4 beta -4) implies 8 alpha -12 beta =28 and 2 alpha - 3 beta =7.`