If the lines x/7 = y/6k = z/7k and x/k = y/1 = z/5 are perpendicular, find the value of k and hence find the equation of the plane containing these lines.

The direction vectors of the lines are given by (7, 6k, 7k) and (k, 1, 5).

Since the two lines are perpendicular, the dot product of the direction vectors must yield zero.

→ 7k + 6k + 35k = 0

Thus, 48k = 0 which implies k = 0. However, if k=0, the lines are not well-defined. Let's re-examine the dot product.

7k + 6k + 35k = 0
48k = 0
k = 0
This result suggests there's an issue with the problem statement as k=0 leads to undefined lines.

Let's assume the question intended for the lines to be skew or intersect and find the plane containing them assuming k is not 0. We will find the plane containing both lines by finding two vectors that lie in that plane and then cross-producting them.

Line 1: x/7 = y/6k = z/7k = λ
Line 2: x/k = y/1 = z/5 = μ

Points on line 1: (7λ, 6kλ, 7kλ)
Points on line 2: (kμ, μ, 5μ)

Let's find a point that lies on both lines. This may not be possible for all values of k.

If λ = μ = 0, then the point (0, 0, 0) is a common point.

Direction vectors:
Line 1: <7, 6k, 7k>
Line 2: <k, 1, 5>

Let's find a vector normal to the plane by taking the cross product of these two direction vectors.

<7, 6k, 7k> x <k, 1, 5> = <30k - 7k, 7k² - 35, 7 - 6k²>
= <23k, 7k² - 35, 7 - 6k²>

The equation of the plane is given by:
23kx + (7k² - 35)y + (7 - 6k²)z = d

Since the point (0,0,0) lies on the plane, d = 0.

Therefore, the equation of the plane is:
23kx + (7k² - 35)y + (7 - 6k²)z = 0

This is the equation of the plane containing the two lines for any k ≠ 0. The value of k can't be determined from the perpendicularity condition as the condition leads to k=0, which renders the lines undefined.