Let the lines
(x-1)/λ=(y-2)/1=(z-3)/2 and
(x+26)/-2=(y+18)/3=(z+28)/λ be coplanar
and P be the plane containing these two lines.
Then which of the following points does NOT lie on P?

Solution

Maths sir · 3 steps

First fix λ using the coplanarity condition, then build the plane and put each point into it.

1. Coplanarity condition. The first line passes through A(1,2,3) with direction (λ,1,2). The second passes through B(-26,-18,-28) with direction (-2,3,λ). They are coplanar when AB⃗ = (-27,-20,-31) and the two directions are linearly dependent, so this determinant must be zero.

   |-27, -20, -31; λ, 1, 2; -2, 3, λ| = 20λ² - 120λ + 180 = 20(λ-3)² = 0

2. Build the plane. So λ = 3, and the directions are (3,1,2) and (-2,3,3). Their cross product is (-3,-13,11), which is the normal to P. Through A(1,2,3) the plane is 3x + 13y - 11z + 4 = 0. Point B also satisfies it, which confirms the plane.

   Point	3x + 13y - 11z + 4	On P?
   (a) (0,-2,-2)	0 - 26 + 22 + 4 = 0	yes
   (b) (-5,0,-1)	-15 + 0 + 11 + 4 = 0	yes
   (c) (3,-1,0)	9 - 13 + 0 + 4 = 0	yes
   (d) (0,4,5)	0 + 52 - 55 + 4 = 1	no

3. So. Only (0,4,5) gives a non-zero value, so it does not lie on P. The answer is (d).

The idea to keep. Two lines are coplanar when the vector joining their points lies in the plane of their directions, so the determinant is zero. The cross product of the directions is then the plane's normal.

Answer from the NTA final answer key. Solution by PrepWiser.