Solution:

Given equation: 3x + 11y = k

We need to find the possible value of x such that the given equation has exactly 3 solutions in which both x and y are positive integers.

To solve this problem, we need to use Diophantine equations.

Diophantine equations are equations that have integer solutions for x and y.

We can rewrite the given equation as:

3x = k - 11y

We can observe that the left-hand side of the equation is a multiple of 3. Therefore, for the right-hand side to be a multiple of 3, k - 11y must be a multiple of 3.

Let k = 3m + r, where m is an integer and r is the remainder when k is divided by 3.

Substituting this in the above equation, we get:

3x = 3m + r - 11y

3(x - m) = r - 11y

We can see that the left-hand side is a multiple of 3. Therefore, for the right-hand side to be a multiple of 3, r - 11y must be a multiple of 3.

We can now list down the possible values of r and the corresponding values of y that make r - 11y a multiple of 3:

r = 0: y = 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99

r = 1: y = 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, 97

r = 2: y = 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98

We need to find the possible value of x such that the given equation has exactly 3 solutions in which both x and y are positive integers.

For x to be a positive integer, we need r - 11y to be positive.

Let's check the possible values of x for each value of r:

r = 0: For y = 1, 4, and 7, r - 11y is positive. Therefore, x = (r - 11y)/3 is a positive integer for these values