To ensure that deadlock never occurs, we need to guarantee that the system can satisfy the resource requests of each process in a safe sequence. The minimum number of resources required can be determined using the Banker's algorithm.

Banker's Algorithm:
1. Available Resources: The number of available resources of each type is denoted by a vector. Let's assume it is Avail[3] = {a, b, c}.
2. Maximum Demand: The maximum number of resources of each type that each process can request is denoted by a matrix. Let's assume it is Max[3][3] = {{x1, y1, z1}, {x2, y2, z2}, {x3, y3, z3}}.
3. Allocated Resources: The number of resources of each type currently allocated to each process is denoted by a matrix. Let's assume it is Alloc[3][3] = {{p1, q1, r1}, {p2, q2, r2}, {p3, q3, r3}}.
4. Need Matrix: The number of resources of each type that each process still needs to complete its execution is denoted by a matrix. It can be calculated as Need[i][j] = Max[i][j] - Alloc[i][j].

In this case:
- Available Resources (Avail) = {3, 4, 5}.
- Maximum Demand (Max) = {{x1, y1, z1}, {x2, y2, z2}, {x3, y3, z3}}.
- Allocated Resources (Alloc) = {{p1, q1, r1}, {p2, q2, r2}, {p3, q3, r3}}.
- Need Matrix (Need) can be calculated as Need[i][j] = Max[i][j] - Alloc[i][j].

Safety Algorithm:
1. Initialize the Work vector as the Available vector and the Finish vector as false for all processes.
2. Find a process Pi whose Finish[i] is false and Need[i] is less than or equal to the Work vector. If no such process exists, go to step 4.
3. Add the Allocated resources of process Pi to the Work vector and set Finish[i] as true. Repeat step 2.
4. If all processes have Finish[i] as true, then the system is in a safe state. Otherwise, it is in an unsafe state.

To ensure that deadlock never occurs, the system must always be in a safe state. In the worst-case scenario, each process may require the maximum number of resources of each type. Hence, to satisfy the maximum demand of each process, the system should have a minimum of:
- x1 + y1 + z1 + x2 + y2 + z2 + x3 + y3 + z3 resources.

In this case, the minimum number of resources required is:
- 3 + 4 + 5 + 3 + 4 + 5 + 3 + 4 + 5 = 36.

Therefore, the correct answer is option 'D' (10).