Given plane, x−2y−z=3 To find the projection let's find the foot of perpendicular from (1,3,, 4) to plane x−2y−z=3
x−1
1
=
y−3
−2
=
z−4
−1
=λ1 (λ1+1)−2(−2λ1+3)−(−λ1+4)=3 ⇒6λ1=12 ⇒ λ1=2 So, foot of perpendicular from (1,3,4) to plane x−2y−z=3 is A (3,−1,2). Let us also find the intersection point of the plane and line
x−1
2
=
y−3
1
=
z−4
2
=λ2 (2λ2+1)−2(λ2+3)−(2λ2+4)=3−2λ2=12 ⇒ λ2=−6 The intersection point of the plane and line is B(−11,−3,−8) Line passing through A and B is
x−3
−14
=
y+1
−2
=
z−2
−10
=μ
x−3
7
=
y+1
1
=
z−2
5
=μ Now, let's find the distance from O(0,0,6) to this line L. Let's say C(7μ+3,μ−1,5μ+2) is any point on L. Then,