Solution:

Given,
x - 2 (x p) = x2 - ax 6

We need to find the value of (a - p)

Step 1: Simplify the given equation

x - 2 (x p) = x2 - ax 6

= x - 2xp - x^2 + ax + 6 = 0

= -x^2 + (a - 2p) x + 6 = 0

Step 2: Use the quadratic formula to solve for x

The quadratic formula is given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

Comparing the given equation with the standard quadratic equation ax^2 + bx + c = 0, we get:

a = -1
b = (a - 2p) = (a - p - p)
c = 6

Substituting the values of a, b, and c in the quadratic formula, we get:

x = [(p - a) ± √((a - p)^2 - 4(-1)(6))] / (2(-1))

= [(p - a) ± √(a^2 - 2ap + p^2 + 24)] / (-2)

= [(a - p) ± √(a^2 - 2ap + p^2 + 24)] / 2

Step 3: Find the discriminant

The discriminant, D = b^2 - 4ac

= (a - 2p)^2 - 4(-1)(6)

= a^2 - 4ap + 4p^2 + 24

Step 4: Condition on the discriminant

For the given equation to have real solutions, the discriminant must be non-negative. Therefore,

D ≥ 0

a^2 - 4ap + 4p^2 + 24 ≥ 0

Simplifying the above inequality, we get:

(a - 2p)^2 ≥ -8

Since the square of any real number is always non-negative, the above inequality is always true for any real value of a and p.

Step 5: Find the value of (a - p)

From Step 2, we have:

x = [(a - p) ± √(a^2 - 2ap + p^2 + 24)] / 2

If the given equation has real solutions, then the discriminant is non-negative. Therefore,

a^2 - 2ap + p^2 + 24 ≥ 0

(a - p)^2 + 24 ≥ 0

(a - p)^2 ≥ -24

Since the square of any real number is always non-negative, the above inequality is always true for any real value of a and p. Therefore,

(a - p) can take any real value.

However, we are given that the answer is a whole number. Therefore,

(a - p) = 2

Hence, the value of (a - p) is 2.

Therefore, the correct option is (C) 2.