Let a = 3i + 2j + xk and b = i - j + k, for some real x. Then |a x b| = r is possible if:

Correct option is D. r ≥ 532
a × b = |i j k
3 2 x
1 -1 1| = (2 + x)i + (x - 3)j - 5k
|a × b| = √((2 + x)² + (x - 3)² + 25) = √(2x² - 2x + 38)
Since |a × b| ≥ 0, 2x² - 2x + 38 ≥ 0. The minimum value occurs at x = -b/2a = 1/2.
Minimum value = 2(1/4) - 2(1/2) + 38 = 37
√37 ≈ 6.08
|a × b| = √(2x² - 2x + 38) ≥ √37 ≈ 6.08
Let's find the minimum value of |a x b|:
|a × b|² = (2 + x)² + (x - 3)² + 25 = 2x² - 2x + 38
To minimize this quadratic, we complete the square:
2(x² - x) + 38 = 2(x - 1/2)² - 1/2 + 38 = 2(x - 1/2)² + 75/2
The minimum occurs at x = 1/2, and the minimum value is √(75/2) = √37.5 ≈ 6.12
Therefore, |a × b| ≥ √37.5 ≈ 6.12
However, there seems to be a calculation error in the provided solution. Let's re-examine:
|a x b| = √((2+x)² + (x-3)² + 25) = √(2x² - 2x + 38)
The discriminant of the quadratic inside the square root is (-2)² - 4(2)(38) = 4 - 304 = -300, which is negative. This means the quadratic is always positive and has a minimum value.
Completing the square:
2x² - 2x + 38 = 2(x² - x) + 38 = 2(x - 1/2)² - 1/2 + 38 = 2(x - 1/2)² + 75/2
The minimum value is √(75/2) = √37.5 ≈ 6.12
The given options do not seem to align with this result. There might be an error in either the question or the provided solution.