Solution:

The given equation is X^2-6kX+9(k^2-k+1)=0.

We need to find the value of k for which both roots of the equation are real, distinct and equal to almost 3.

Condition for real and distinct roots:

For real and distinct roots, the discriminant of the equation should be greater than 0. The discriminant of the given equation is:

D = b^2 - 4ac

Here, a=1, b=-6k and c=9(k^2-k+1)

D = (-6k)^2 - 4(1)(9(k^2-k+1))

D = 36k^2 - 36(k^2-k+1)

D = 72k - 36

D = 36(2k - 1)

For real and distinct roots, D > 0

36(2k - 1) > 0

2k - 1 > 0

k > 1/2

Condition for roots equal to almost 3:

We are given that both roots are equal to almost 3. Let the roots be p and q.

p + q = 6k/1 = 6k

pq = 9(k^2-k+1)/1 = 9k^2 - 9k + 9

Since both roots are almost 3, we have:

p = q = 3 + a (where a is a small quantity)

p + q = 6 + 2a

pq = (3+a)^2 = 9 + 6a + a^2

Substituting these values in the above equations, we get:

6 + 2a = 6k

9 + 6a + a^2 = 9k^2 - 9k + 9

Simplifying these equations, we get:

a = 3k - 3

a^2 - 3a + 3 = 3k^2 - 3k

Substituting the value of a, we get:

(3k-3)^2 - 3(3k-3) + 3 = 3k^2 - 3k

Simplifying this equation, we get:

9k^2 - 33k + 27 = 0

Dividing both sides by 3, we get:

3k^2 - 11k + 9 = 0

Solving this quadratic equation, we get:

k = 1 or k = 3/2

But k > 1/2 (from the condition for real and distinct roots)

Therefore, the only value of k for which both roots of the equation are real, distinct and equal to almost 3 is:

k = 3/2

Hence, the number of values of k is 1.

Take D=0