The set of all real values of λ for which exactly two common tangents can be drawn to the circles x² + y² - 8x - 8y + 6 = 0 and x² + y² - 12x - 12y + λ = 0 is the interval:

Two circles can have exactly two common tangents only if the circles intersect each other. For the circles to intersect each other, the following conditions must hold true:

r₁ + r₂ > dist(c₁, c₂)
r₂ - r₁ < dist(c₁, c₂)

The given equations of circles are:
x² + y² - 8x - 8y + 6 = 0
x² + y² - 12x - 12y + λ = 0

The centers and radii of the circles are:
Circle 1: Center (4, 4), Radius r₁ = √(4² + 4² - 6) = √22
Circle 2: Center (6, 6), Radius r₂ = √(6² + 6² - λ) = √(72 - λ)

Distance between centers = √((6 - 4)² + (6 - 4)²) = √8 = 2√2

Applying the conditions for intersection:
√22 + √(72 - λ) > 2√2
√(72 - λ) - √22 < 2√2

Solving the inequalities:
√(72 - λ) > 2√2 - √22 ≈ -1.1
√(72 - λ) < 2√2 + √22 ≈ 6.4

Squaring both sides:
72 - λ > 0 => λ < 72
72 - λ < 41 => λ > 31

Therefore, 31 < λ < 72
However, this is not one of the options.
Let's reconsider the condition for exactly two common tangents. This occurs when the distance between the centers is between the difference and sum of the radii. That is:
|r₁ - r₂| < dist(c₁, c₂) < r₁ + r₂
|√22 - √(72 - λ)| < 2√2 < √22 + √(72 - λ)

Solving for the right inequality:
2√2 < √22 + √(72 - λ)
√(72 - λ) > 2√2 - √22 ≈ -1.1 (always true since √(72-λ) must be positive)

Solving for the left inequality:
|√22 - √(72 - λ)| < 2√2
-2√2 < √22 - √(72 - λ) < 2√2
-2√2 - √22 < -√(72 - λ) < 2√2 - √22
2√2 + √22 > √(72 - λ) > √22 - 2√2
(2√2 + √22)² > 72 - λ > (√22 - 2√2)²
72 - λ < (2√2 + √22)² ≈ 41
72 - λ > (√22 - 2√2)² ≈ 0
λ > 31
λ < 72
Combining these, we have 31 < λ < 72. Let's examine the options given: (12, 24), (18, 42), (12, 32), (18, 48). Only (18, 42) falls within the derived range. There must be a calculation error in the provided solution. The correct interval should be (31, 72), however this is not among the options provided. There might be an error in the question or provided solution.