Three Dimensional Geometry

Normal of plane = |î? k|
|1 0|
|0 1 -1|
n = -î+? +k
D.R.'s = -1,1,1
Plane => -1 (x-1)+1 (y-0)+1 (z-0) = 0
=> x-y-z-1=0
If (x, y, z) is foot of perpendicular of M (1,0,1) on the plane then
(x-1)/-1 = (y-0)/1 = (z-1)/1 = - (1-0-1-1)/3
x=4/3, y=-1/3, z=2/3
α+β+γ = 4/3 - 1/3 + 2/3 = 5/3

Mid point P = (3,1,1)
Normal of plane is along the line AB.
D.R.'s of normal = 4-2, -2-4, 3-1 (-1) = 2, -6, 4
= 1, -3, 2
Plane -> 1 (x-3) - 3 (y-1) + 2 (z-1) = 0
=> x - 3y + 2z - 2 = 0

Equation of
AB = r = (î + j) + λ (3j - 3k)
Let coordinates of M
= (1, (1 + 3λ), -3λ).
PM = -3î + (3λ - 1)j - 3 (λ + 1)k
AB = 3j - 3k
? PM ⊥ AB ⇒ PM · AB = 0
⇒ 3 (3λ - 1) + 9 (λ + 1) = 0
⇒ λ = -1/3

∴ M = (1,0,1)
Clearly M lies on 2x + y - z = 1.

r = î (1 + 12l) + j (-1) + k (l)
r = î (2 + m) + j (m - 1) + k (-m)
For intersection
1 + 2l = 2 + m
-1 = m - 1
l = -m
from (ii) m = 0
from (iii) l = 0
These values of m and l do not satisfy equation (1).
Hence the two lines do not intersect for any values of l and m.