What will be the output when the following code is executed: const foo = () => console.log('First'); const bar = () => setTimeout(() => console.log('Second')); const baz = () => console.log('Third'); bar(); foo(); baz();?

The code provided defines three functions: foo, bar, and baz. The execution flow of these functions reveals the output order.

When bar() is called, it invokes a setTimeout function that schedules the output of 'Second' to occur after the timer (default is 0 milliseconds) has elapsed. This means that while 'Second' is scheduled, control is immediately returned to the next line of code since it's asynchronous.

Next, foo() is executed, which immediately outputs 'First' to the console. After that, baz() is called, which outputs 'Third' to the console as well.

At the end of this sequence, the scheduled function from bar will execute and output 'Second' to the console, but only after the synchronous operations (the calls to foo and baz) have completed.

So, in the order of execution:

1. bar() schedules 'Second' (but it doesn't log it immediately).
2. foo() logs 'First'.
3. baz() logs 'Third'.
4. After all synchronous code has executed, 'Second' is logged by the scheduled callback from bar.

Hence, the final order of outputs is 'First