If lines x/2 = (y+1)/3 = z/4 and x/3 = (y-k)/2 = z/1 intersect, then find the value of k and hence find the equation of the plane containing these lines.

Any point on x/2 = (y+1)/3 = z/4 = λ is (2λ, -1+3λ, 4λ)
Any point on x/3 = (y-k)/2 = z/1 = μ is (3μ, k+2μ, μ).
For the two lines to intersect, the following three equations must be satisfied simultaneously:
2λ = 3μ ⇒ 2λ - 3μ = 0
-1 + 3λ = k + 2μ ⇒ 3λ - 2μ = k + 1
4λ = μ ⇒ 4λ - μ = 0
Solving the above equations, we get, λ = 0, μ = 0, k = -1
The point of intersection is therefore, (0, -1, 0).
The normal to the plane containing the lines is obtained by the cross product of the vectors:
(2i + 3j + 4k) × (3i + 2j + k) = -5i + 10j - 5k.
Since the plane contains both the lines it must contain the point of intersection as well, so in order to find the constant d in the equation of the plane →r.→n = d, where →r is any point on the plane and →n is the normal vector to the plane, we substitute the point of intersection for →r in the equation:
d = (0i - 1j + 0k) (-5i + 10j - 5k) = -10.
Therefore, the equation of the plane is -5x + 10y - 5z = -10 or x - 2y + z = 2