vSAN will use a 4+1 RAID-5 data placement scheme with parity will be used is the correct answer because vSAN 8 ESA uses adaptive RAID-5 erasure coding that depends on the number of hosts in the cluster. If the cluster has 6 or more hosts, vSAN will use a 4+1 RAID-5 scheme, where the data is written as a stripe of 4 data bits and 1 parity bit across 5 hosts. This provides a failure tolerance of 1 (FTT=1) and a space efficiency of 1.25x. If the cluster has less than 6 hosts (3 to 5), vSAN will use a 2+1 RAID-5 scheme, where the data is written as a stripe of 2 data bits and 1 parity bit across 3 hosts. This also provides a failure tolerance of 1 (FTT=1) but a space efficiency of 1.5x. In this case, the cluster has 5 hosts, so vSAN will use the 4+1 RAID-5 scheme.

The other options are incorrect for the following reasons:

A, vSAN will use a 2+1 RAID-5 data placement scheme with parity will be used, is incorrect because vSAN will only use this scheme if the cluster has less than 6 hosts but more than 2 hosts. In this case, the cluster has 5 hosts, so vSAN will use the 4+1 RAID-5 scheme.

B, RAID 5 will provide an FTT=2 level of protection in this case, is incorrect because RAID 5 can only provide an FTT=1 level of protection, regardless of the number of hosts or the data placement scheme. To achieve an FTT=2 level of protection, vSAN would need to use RAID 6 erasure coding, which requires at least 6 hosts in the cluster.

D, vSAN will spread the components across all of the disk groups, is incorrect because vSAN will not necessarily spread the components across all of the disk groups in the cluster. vSAN will only spread the components across as many disk groups as needed to meet the storage policy requirements and to balance the load and capacity. In this case, vSAN will only need to spread the components across 5 disk groups for each stripe of RAID-5 data. References:

VMware vSAN Specialist v2 Exam Preparation Guide, page 11

Adaptive RAID-5 Erasure Coding with the Express Storage Architecture in vSAN 8