Answer :

Given the equation of line are-

∴ any random point on this line is given by (2λ+1, 3λ-1, 4λ+1)

Another line is:

∴ any random point on this line is given by (μ+3, 2μ+k, μ)

At point of intersection of the lines the random coordinates must coincide.

∴ 2λ + 1 = μ+3

⇒ 2λ - μ = 2 …(1)

Also,

4λ + 1 = μ

⇒ μ = 4λ + 1 …(2)

Adding equation 1 and 2,we get-

2λ = 4λ + 3

⇒

⇒

As, 3λ – 1 = 2μ + k

∵ we have the values of λ and μ.

∴ k = 3λ – 2μ – 1

⇒ k = 3(-3/2) – 2(-5) – 1

⇒ k = 9/2

Now we need to find the equation of the plane containing these 2 lines.

For this we need the normal vector to plain and a point on plane.

For normal we need to take the cross product of direction ratios of line.

Direction ratio of line 1 is (2,3,4)

And direction ratio of line 2 is ( 1,2,1)

∴ the normal vector is given as -

∴

⇒

Let be any random vector on pthe lane.

The equation of plane is given by –

where a is any defined vector on planthe e.

As lines lie on a plane, so one of its points can be taken as a poi t on the plane.

∴ (2λ+1, 3λ-1, 4λ+1) will give a point on putting λ = 0

Point is (1, -1, 1)

∴

Hence, the equation of the plane is:

⇒ -5x + 2y + z = -5-2+1 = -6

∴ Equation is: -5x+2y+z = -6

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