Let the foot of perpendicular from the point A 4 , 3 , 1 on the plane P : x - y + 2 z + 3 = 0 be N . If B ( 5 , α , β ) , α , β ∈ ℤ is a point on plane P such that the area of the triangle A B N is 3 2 , then α 2 + β 2 + α β is equal to _____________

Step-by-step solution

Given, The foot of perpendicular from the point A 4 , 3 , 1 on the plane P :   x - y + 2 z + 3 = 0 be N So, by foot of perpendicular formula we get, x 1 - 4 1 = y 1 - 3 - 1 = z 1 - 1 2 = - 4 - 3 + 2 + 3 1 2 + 1 2 + 2 2 ⇒ N ≡ x 1 , y 1 , z 1 = ( 3 , 4 , - 1 ) And B 5 , α , β , α , β ∈ ℤ is a point on plane P such that the area of the triangle A B N is 3 2 , So putting the value of point B 5 , α , β in plane we get, 5 - α + 2 β + 3 = 0 ⇒ &#945