See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Plane containing the line will be given as (4x – 2y – 10) +  (5y – 4y + 3) = 0 4x (5 2)y 4 z 3 10 0         the plane passing through (4, 3, 7) will give  16x (5 2)y 4 z 3 10 0         10 0 0    i.e., 4x – 2y – 10 = 0 If this rotated an angle 2  the plane will be 4x + (5’ –2)y –4’z + (2’ – 10) = 0 Also 16 (5 ' 2)( 2) 0      ' 2; i.e. 4x 8y 8z 4     x y z OA yOB zOC 0 l m n          