We are given with lines,

Let these lines be L₁ and L₂, such that

Where λ, μ ∈ ℝ

We need to show that lines L₁ and L₂ intersect.

In order to show this, let us find out any point on line L₁ and line L₂.

For line L₁:

We need to find the values of x, y and z. So,

Take .

⇒ x – 1 = 2λ

⇒ x = 2λ + 1

Take .

⇒ y – 2 = 3λ

⇒ y = 3λ + 2

Take .

⇒ z – 3 = 4λ

⇒ z = 4λ + 3 …(i)

∴, any point on line L₁ is (2λ + 1, 3λ + 2, 4λ + 3).

For line L₂:

We need to find the values of x, y and z. So,

Take .

⇒ x – 4 = 5μ

⇒ x = 5μ + 4

Take .

⇒ y – 1 = 2μ

⇒ y = 2μ + 1

Take

⇒ z = μ …(ii)

∴, any point on line L₂ is (5μ + 4, 2μ + 1, μ).

Since, if line L₁ and L₂ intersects then there exists λ and μ such that,

(2λ + 1, 3λ + 2, 4λ + 3) ≡ (5μ + 4, 2μ + 1, μ)

⇒ 2λ + 1 = 5μ + 4 …(iii)

3λ + 2 = 2μ + 1 …(iv)

4λ + 3 = μ …(v)

Substituting value of μ from equation (v) in equation (iv), we get

3λ + 2 = 2(4λ + 3) + 1

⇒ 3λ + 2 = 8λ + 6 + 1

⇒ 3λ + 2 = 8λ + 7

⇒ 8λ – 3λ = 2 – 7

⇒ 5λ = -5

⇒ λ = -1

Putting λ = -1 in equation (v), we get

4(-1) + 3 = μ

⇒ μ = -4 + 3

⇒ μ = -1

To check, put λ = -1 and μ = -1 in equation (iii),

2(-1) + 1 = 5(-1) + 4

⇒ -2 + 1 = -5 + 4

⇒ -1 = -1

∴, λ and μ also satisfy equation (iii).

So, z-coordinate from equation (i),

z = 4λ + 3

⇒ z = 4(-1) + 3 [∵, λ = -1]

⇒ z = -4 + 3

⇒ z = -1

And z-coordinate from equation (ii),

z = μ

⇒ z = -1 [∵, μ = -1]

So, the lines intersect.

Their point of intersection is (5μ + 4, 2μ + 1, μ) = (5(-1) + 4, 2(-1) + 1, -1)

Or (5μ + 4, 2μ + 1, μ) = (-5 + 4, -2 + 1, -1)

Or (5μ + 4, 2μ + 1, μ) = (-1, -1, -1)

Thus, the given lines intersect, and the point of intersection is (-1, -1, -1).