2
3
1
4
Lines are coplanar ⇔ |1 1 -3λ/2| = 0
|1 λ²/2 -2| = 0
|0 -λ²/2 +1|
⇒ |20 4 -12λ/2 1λ²/2 -2 0 -λ²/2 1| = 0
⇒ 2(4 - λ²/2) + 4(1 + 3λ²/2) = 0
⇒ 8 - λ² + 4 + 6λ² = 0
⇒ 5λ² = -12
⇒ λ² = -12/5 which is not possible since λ is real.
Let the lines be
L1: x-1/1 = y-2/1 = z+3λ/2
L2: x/1 = y-λ²/2 = z-2/1
For coplanarity
|1 1 -3λ/2| = 0
|1 λ²/2 -2| = 0
|0 -λ²/2 1|
2(4-λ²/2) + 4(1 + 3λ²/2) = 0
8 - λ² + 4 + 6λ² = 0
5λ² = -12
This is not possible since λ is real.
If lines are coplanar then
|a1 - a2 b1 - b2 c1 - c2| = 0
|a1 b1 c1| = 0
|a2 b2 c2|
2(4 - λ⁴) + 4(λ² + 3λ²) = 0
8 - 2λ⁴ + 16λ² = 0
λ⁴ - 8λ² - 4 = 0
λ² = 8 ± √64 + 16/2 = 4 ± 2√5
λ² = 4 + 2√5 or 4 - 2√5
Since λ is real, there are four values of λ