Given information:
Family of lines: a(3x + 4y + 6) + b(x + y + 2) = 0
Point: P(2, 3)

To find the line of the family that is situated at the greatest distance from point P(2, 3), we need to find the line that maximizes the perpendicular distance between the line and point P.

Let's find the perpendicular distance between a line and a point using the formula:

Perpendicular distance (d) = |a*x1 + b*y1 + c| / sqrt(a^2 + b^2)

where a, b, c are the coefficients of the line equation and (x1, y1) is the coordinates of the point.

Steps to find the line:
1. Convert the equation of the family of lines to the standard form of a line.
2. Calculate the perpendicular distance between each line and point P.
3. Find the line that gives the maximum perpendicular distance.

1. Convert the equation of the family of lines to the standard form of a line:
a(3x + 4y + 6) + b(x + y + 2) = 0

Distribute the coefficients:
(3a + b)x + (4a + b)y + (6a + 2b) = 0

Comparing with the standard form of a line, we get:
A = 3a + b
B = 4a + b
C = 6a + 2b

2. Calculate the perpendicular distance between each line and point P:
Using the formula for perpendicular distance, we have:
d = |A*x1 + B*y1 + C| / sqrt(A^2 + B^2)

Substituting the values of A, B, C, x1, and y1, we get:
d = |(3a + b)*2 + (4a + b)*3 + (6a + 2b)| / sqrt((3a + b)^2 + (4a + b)^2)

Simplifying the expression, we have:
d = |12a + 7b| / sqrt(9a^2 + 10ab + b^2)

3. Find the line that gives the maximum perpendicular distance:
To find the line that maximizes the perpendicular distance, we need to maximize the numerator and minimize the denominator of the expression.

The maximum value of |12a + 7b| is obtained when a = 0 and b = 1.
Substituting these values back into the equation, we get:
d = |7| / sqrt(1) = 7

Therefore, the line of the family that is situated at the greatest distance from point P(2, 3) has a perpendicular distance of 7 units.

Conclusion:
The line of the family situated at the greatest distance from the point P(2, 3) has the equation:
4x - 3y + 8 = 0, which is option A.