Given:
System of linear equations:
1) (3k + 1)x + 3y - 2 = 0
2) (k^2 + 1)x + (k - 2)y - 5 = 0

To find:
The value of k for which the system of linear equations has no solution.

Solution:
To determine if the system of linear equations has a solution, we need to check if the two equations are consistent or inconsistent.

Consistency of a System of Linear Equations:
A system of linear equations is consistent if it has at least one solution. It can have either a unique solution or infinitely many solutions.

Inconsistency of a System of Linear Equations:
A system of linear equations is inconsistent if it has no solution. This occurs when the equations are parallel and never intersect.

Expressing equations in slope-intercept form:
To analyze the system of equations, let's express the equations in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.

1) (3k + 1)x + 3y - 2 = 0
3y = -(3k + 1)x + 2
y = -(3k + 1)/3x + 2/3

2) (k^2 + 1)x + (k - 2)y - 5 = 0
(k - 2)y = -(k^2 + 1)x + 5
y = -(k^2 + 1)/(k - 2)x + 5/(k - 2)

Comparing the slopes:
For the two lines to be parallel (and have no solution), their slopes must be equal.

Comparing the slopes of the two equations:
-(3k + 1)/3 = -(k^2 + 1)/(k - 2)

Case 1: k - 2 ≠ 0
If k - 2 ≠ 0, we can multiply both sides of the equation by 3(k - 2) to eliminate the denominators:

-(3k + 1)(k - 2) = -3(k^2 + 1)

Simplifying the equation:
-3k^2 + 5k + 2 = -3k^2 - 3

Cancelling out the -3k^2 terms:
5k + 2 = -3

Simplifying further:
5k = -5
k = -1

Case 2: k - 2 = 0
If k - 2 = 0, then k = 2. However, we need to exclude this value as it would result in division by zero in the equation for the slope.

Conclusion:
From the two cases, the only value of k that results in no solution for the system of linear equations is k = -1. Therefore, the correct answer is option 'B'.