Question

Consider the lines $L_1: \frac{x+1}{3} = \frac{y+2}{1} = \frac{z+1}{2}$ $L_2: \frac{x-2}{1} = \frac{y+2}{2} = \frac{z-3}{3}$ The distance of the point (1,1,1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_1$ and $L_2$ is $\frac{10 + \lambda}{\sqrt{75}}$ then $\lambda =$

Answer 1

3

Answer 2

Explanation of the solution:

1. The direction ratios of L₁ and L₂ are (3, 1, 2) and (1, 2, 3) respectively.
2. The normal of the plane is perpendicular to both these vectors, so take the cross product:
     n = (3, 1, 2) × (1, 2, 3) = (1·3 – 2·2, –(3·3 – 2·1), 3·2 – 1·1) = (3 – 4, –(9 – 2), 6 – 1) = (–1, –7, 5).
3. The plane passes through the point (–1, –2, –1), so its equation is:
     –1(x + 1) – 7(y + 2) + 5(z + 1) = 0.
4. The distance from the point (1, 1, 1) to the plane is given by:
     Distance = |–1(1+1) – 7(1+2) + 5(1+1)| / √(1²+7²+5²)
          = |–2 – 21 + 10| / √75
          = |–13| / √75 = 13/√75.
5. Given that the distance is (10 + λ)/√75, equate numerators:
     13 = 10 + λ ⟹ λ = 3.

Answer:
λ = 3