We are given the equation of a line,

We need to find equations of the two lines through the origin which intersect the given line.

We know the theorem that says, equation of a line with direction ratios d = (b₁, b₂, b₃) that passes through the point (x₁, y₁, z₁) is given by the formula,

Also, we know that

Angles between two lines whose direction ratio are d₁ and d₂ is θ given by,

Using these theorems, we will find equation of the two lines.

Let the equation of a line be,

And it given that the lines pass through the origin.

⇒ (a₁, a₂, a₃) ≡ (0, 0, 0)

So, equation of both lines that passes through the origin will be of the form:

…(i)

Let

…(ii)

⇒ Direction ratio of the line = (2, 1, 1)

⇒ d₁ = (2, 1, 1) …(iii)

Where, d₁ = Direction ratio of the line (ii).

Let us represent the direction ratio in position vector.

So,

…(iv)

Any point of on the given line is given by (x, y, z). So, from (ii)

Take .

⇒ x – 3 = 2μ

⇒ x = 2μ + 3

Take .

⇒ y – 3 = μ

⇒ y = μ + 3

Take .

⇒ z = μ

So, any point on the line (ii) is

P(2μ + 3, μ + 3, μ)

Since, line (i) passes through origin, we can say that

(b₁, b₂, b₃) ≡ (2μ + 3, μ + 3, μ)

⇒ Direction ratio of line (i) = (2μ + 3, μ + 3, μ)

⇒ d₂ = (2μ + 3, μ + 3, μ) …(v)

Representing the direction ratio in position vector.

So,

…(vi)

We know, by theorem

Substituting the values of d₁ and d₂ from (iv) and (vi) in the above equation. Also, put θ = π/3 as mentioned in the question.

Solving numerator,

Solving denominator,

And,

Putting the values, we get

By cross-multiplying,

⇒ 6√(μ² + 3μ + 3) = 2(6μ + 9)

⇒ 6√(μ² + 3μ + 3) = 2 × 3(2μ + 3)

⇒ 6√(μ² + 3μ + 3) = 6(2μ + 3)

⇒ √(μ² + 3μ + 3) = 2μ + 3

Taking square on both sides,

⇒ (√(μ² + 3μ + 3))² = (2μ + 3)²

⇒ μ² + 3μ + 3 = (2μ)² + (3)² + 2(2μ)(3) [∵, (a + b)² = a² + b² + 2ab]

⇒ μ² + 3μ + 3 = 4μ² + 9 + 12μ

⇒ 4μ² – μ² + 12μ – 3μ + 9 – 3 = 0

⇒ 3μ²+ 9μ + 6 = 0

⇒ 3(μ² + 3μ + 2) = 0

⇒ μ² + 3μ + 2 = 0

⇒ μ² + 2μ + μ + 2 = 0

⇒ μ(μ + 2) + (μ + 2) = 0

⇒ (μ + 1)(μ + 2) = 0

⇒ (μ + 1) = 0 or (μ + 2) = 0

⇒ μ = -1 or μ = -2

So,

From (v):

Direction Ratio = (2μ + 3, μ + 3, μ)

Put μ = -1,

Direction Ratio = (2(-1) + 3, (-1) + 3, -1)

⇒ Direction Ratio = (-2 + 3, -1 + 3, -1)

⇒ Direction Ratio = (1, 2, -1) …(vi)

Now, put μ = -2,

Direction Ratio = (2(-2) + 3, (-2) + 3, -2)

⇒ Direction Ratio = (-4 + 3, -2 + 3, -2)

⇒ Direction Ratio = (-1, 1, -2) …(vii)

Using direction ratios in (vi) and (vii), in equation (i):

And

Thus, the required two lines are and .