Given information:
- The sum of length, breadth, and height of a cuboidal box is 19 cm.
- The length of the diagonal is 11 cm.

Let's assume:
- The length of the box is L cm.
- The breadth of the box is B cm.
- The height of the box is H cm.

Using the given information, we can form two equations:
1) L + B + H = 19 (sum of length, breadth, and height)
2) Diagonal = 11

To find the surface area of the cuboidal box, we need to find the length, breadth, and height of the box.

To find the length, breadth, and height, we can use the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse (diagonal) is equal to the sum of the squares of the other two sides.

Let's assume:
- The diagonal of the cuboidal box is d cm.
- The length of the cuboidal box is l cm.
- The breadth of the cuboidal box is b cm.
- The height of the cuboidal box is h cm.

Using the Pythagorean theorem, we can form the following equation:
d^2 = l^2 + b^2 + h^2

Since we are given that the diagonal is 11 cm, we can substitute the values and simplify the equation:
11^2 = l^2 + b^2 + h^2
121 = l^2 + b^2 + h^2 ...(Equation 1)

Now, using the sum of length, breadth, and height, we have the equation:
l + b + h = 19 ...(Equation 2)

Solving Equations 1 and 2 simultaneously will give us the values of l, b, and h.

By substituting the value of h from Equation 2 into Equation 1, we get:
121 = l^2 + b^2 + (19 - l - b)^2

Expanding and simplifying the equation:
121 = l^2 + b^2 + (361 - 38l - 38b + l^2 + b^2 - 2lb)

Combining like terms:
121 = 2l^2 + 2b^2 - 38l - 38b + 361

Rearranging the equation:
2l^2 + 2b^2 - 38l - 38b + 240 = 0

Dividing the entire equation by 2:
l^2 + b^2 - 19l - 19b + 120 = 0

We have a quadratic equation in the form of Ax^2 + Bx + C = 0, where A = 1, B = -19, and C = 120.

To solve this quadratic equation, we can either factorize it or use the quadratic formula.

Using the quadratic formula:
l = (-B + √(B^2 - 4AC)) / (2A)
l = (-(-19) + √((-19)^2 - 4(1)(120))) / (2(1))
l = (19 + √(361 - 480)) / 2
l = (19 + √(-119)) / 2

Since the area of a