If the angle between the lines x/2 = y/2 = z/1 and (5-x)/2 = (7y+14)/p = (z-3)/4 is cos⁻¹(2/3), then p is equal to?

Let the direction ratios of the first line be a1 = 2, b1 = 2, c1 = 1.
Let the direction ratios of the second line be a2 = -2, b2 = p/7, c2 = 4.
The angle θ between the two lines is given by:
cos θ = (a1a2 + b1b2 + c1c2) / (√(a1² + b1² + c1²) * √(a2² + b2² + c2²))
Given that cos θ = 2/3
Substituting the values:
2/3 = (2(-2) + 2(p/7) + 1(4)) / (√(2² + 2² + 1²) * √((-2)² + (p/7)² + 4²))
2/3 = (-4 + 2p/7 + 4) / (√9 * √(4 + p²/49 + 16))
2/3 = (2p/7) / (3 * √(20 + p²/49))
2/3 = (2p/21) / √(20 + p²/49)
2 * 21√(20 + p²/49) = 6p
7√(20 + p²/49) = p
Squaring both sides:
49(20 + p²/49) = p²
980 + p² = p²
This equation is incorrect. There must be a mistake in the direction ratios of the second line. Let's reconsider the second line's equation:
(5-x)/2 = (7y+14)/p = (z-3)/4
The direction ratios are a2 = -2, b2 = p/7, c2 = 4
cos θ = (a1a2 + b1b2 + c1c2) / (√(a1² + b1² + c1²) * √(a2² + b2² + c2²))
2/3 = ((2)(-2) + (2)(p/7) + (1)(4)) / (√(4+4+1)√(4 + p²/49 + 16))
2/3 = (-4 + 2p/7 + 4) / (3√(20 + p²/49))
2/3 = (2p/7) / (3√(20 + p²/49))
2(3√(20 + p²/49)) = 6p/7
7√(20 + p²/49) = p
Squaring both sides:
49(20 + p²/49) = p²
980 + p² = p²
This is still incorrect. Let's correct the direction ratios for the second line. The equation should be:
(5-x)/2 = (7y+14)/p = (z-3)/4
Direction ratios are: a2 = -2, b2 = p/7, c2 = 4
cos θ = (a1a2 + b1b2 + c1c2) / (√(a1² + b1² + c1²) * √(a2² + b2² + c2²))
2/3 = (-4 + 2p/7 + 4) / (3√(20 + p²/49))
2/3 = (2p/7) / (3√(20 + p²/49))
14√(20 + p²/49) = 2p
7√(20 + p²/49) = p
49(20 + p²/49) = p²
980 + p² = p² This is still incorrect. There seems to be an error in the problem statement or the given options.