CrowdStrike CCSE-204 Dumps Questions Answers

The correct answer is A. Regenerate a new API key directly from the platform .

CrowdStrike guidance around connector onboarding shows that after a connector is created, you generate an API key in the platform and use that key for the integration. Related integration guidance also shows a Regenerate API key action in the platform flow, which is the correct response when a key may be exposed or compromised.

Why the other options are incorrect:

B does not address credential compromise; recreating the connector event does not invalidate the exposed key.

C is incorrect because the issue is not viewing or cloning details; the security action is to rotate/regenerate the credential.

D is incorrect because CrowdStrike documentation consistently indicates secrets/keys are generated in-platform and may only be shown once, meaning Support is not the normal mechanism to retrieve and resend an existing secret.

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The correct answer is C. logscale-collector .

CrowdStrike’s Falcon LogScale Collector installation documentation states that the service name varies by installation method. It explicitly says that for Full Installation the service is called logscale-collector , while Custom Installation uses humio-log-collector . Since the question specifically refers to deployment using the Fleet Management interface commands , that aligns with the Full Installation workflow, so the correct service name is logscale-collector .

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CrowdStrike LogScale documentation states that groupBy() is used to group events by one or more specified fields, similar to SQL GROUP BY. The documentation also says the function parameter accepts aggregate functions, and its default is count(as=_count). That means the query that explicitly groups by ComputerName, UserName, and CommandLine and applies function=count() is the correct way to output the frequency of each unique combination of those three fields.

Why the other options are incorrect:

A is incorrect because table() formats output rows but does not aggregate unique combinations into frequencies the way groupBy() does. Adding count() after table() does not produce grouped counts for each unique triplet. B is incorrect because table() is not the aggregation function documented for grouped frequency counting; groupBy() is. D is close, but it relies on the default count behavior rather than explicitly specifying function=count(). Since the question asks which query will output the frequency of a unique set, C is the most correct and explicit choice.