For divisibility by $2$, the sequence is $$2,4,6,8,.....\ 1000000$$
For divisibility by $3$, $$3,6,9,12,.....\ 999999$$
For divisibility by $19$, $$19,38,57,76,.....\ 999989$$
But if we add all of these, we will be counting some numbers like $6, 38,57$ more than once. So we have to use inclusion-exclusion principle.
For divisibility by $6$, the sequence is $$6, 12, 18, 24,......\ 999996$$
For divisibility by $38$, $$38,76,114,.....\ 999970$$
For divisibility by $57$, $$57,114,.....\ 999951$$
So now we can subtract these sequences from the first three.
But then we will be subtracting multiples of $114$ once more than necessary. So we need to add in the sequence of multiples of $114$, $$114,228,.....999894$$
Now you can find the cardinality of each of these and add/subtract according to the inclusion-exclusion principle.
EDIT according to comment:
The sequence of numbers divisible by $3$ is an arithmetic progression (with common difference $3$ and first term $3$). Using the formula for general term, $$a_n=a+(n-1)d$$ $$999999=3+(n-1)3$$ $$n=333333$$ So the cardinality of this set is $333333$. This method can be used for any such sequence.
An alternative method is as follows: If we divide all of the terms by a specific number, the number of terms remains the same. So $3,6,9,12,.....\ 999999$ has the same number of terms as $1,2,3,4......\ 333333$. It's obvious that the number of terms is $333333$. You can use this method for all such sequences.
Now to clear your second doubt, those numbers will have to be excluded which appear in more than one of the $3$ sets. For example, which numbers will be common to the first two sets? The first one has multiples of $2$, and the second one has multiples of $3$. So naturally, the common numbers are the ones that are multiples of both $2$ and $3$, i.e. they are multiples of $6$.
Similarly, the numbers common to the first and third sets are multiples of $38$, the numbers common to the second and third sets are multiples of $57$, and the numbers common to all three sets are multiples of $114$. I have already mentioned above how to calculate the cardinality of these.
Now, to make you understand which sets to include and exclude, I'm going to take a simpler example. Please note that it is not the same question as above. Consider the numbers from $1$ to $30$. If you want to find numbers that are multiples of $2$, $3$ and $5$, the three sets are as follows-
We see that the first set has $15$ elements, the second has $10$ and the third has $6$.
The sets of common numbers to each pair of sets are (multiples of $6$, $10$ and $15$ respectively): $$6,12,18,24,30$$ $$10,20,30$$ $$15,30$$
We see that exactly one number is common to these three sets: $$30$$
Now: if we count only the first three sets, we end up with cardinality of $$15+10+6=31$$
We see that numbers like $6, 10, 15$ etc. are being counted twice. So we subtract the cardinalities of these three.
But wait! We've counted $30$ thrice, and subtracted it thrice too. So we need to count it once more.
We can confirm that the final set that we need is (by counting manually) $$2,3,4,5,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30$$
Which has a cardinality of $22$, so our answer is correct.
Now we can extend this principle to the question and solve it.