Ask the following question of all three guards:

If I asked you which door you were guarding, would you say it was the door to heaven?

Now the number of Yeses (Y) will be between 0 and 3 inclusive.

If Y=1, go through that door. The position may either be

1.

(Michael, John, Vlad) = (Yes, No, No)

in which case you go to heaven, or it may be one of

(No, Yes, No) and(No, No, Yes)

in which case you go to hell.

If Y=2, namely

3.

(Yes, Yes, No), (Yes, No, Yes), or (No, Yes, Yes).

then pick one of the Yeses at random and ask the utterer the same question again. If he changes to a No,

go through the door which had the other Yes.

If he doesn’t,

pick one of the Yes doors at random and go through it.

If Y=3, namely

(Yes, Yes, Yes)

then again, pick one of the Yeses at random and ask the utterer the same question again.

The number of remaining Yeses will be two or three. Pick one of them at random and go through the corresponding door.

The last possibility is that Y=0:

5.

(No, No, No)

Oh dear.

Pick a guard at random and repeat the question. The number of Yeses will now be one or zero. If it is one, go through the corresponding door. If it is zero, pick one of three doors at random and hope for the best.

This is ignoring the magic stone. I think StephenTG got the right answer. This was basically a trick question. But if you ignore the magic stone, the resulting question is still a good puzzle, and the above may be the best strategy.

Edit

I’ve edited this in light of Amorydai’s helpful comment.

For the three equiprobable cases of which door leads to heaven, the probabilities of each guard saying “Yes” are as follows (writing M, J, V for Michael, John, Vlad):

Case 1: M’s door leads to heaven
M: $0.75^2 + 0.25^2 = 0.625$;
J: $2 * 0.7*0.3 = 0.42$
V: $2 * 0.9*0.1 = 0.18$

Case 2: J’s door leads to heaven
M: $2 * 0.75 * 0.25 = 0.375$;
J: $0.7^2 * 0.3^2 = 0.58$
V: $2 * 0.9*0.1 = 0.18$

Case 3: V’s door leads to heaven
M: $2 * 0.75 * 0.25 = 0.375$;
J: $2 * 0.7*0.3 = 0.42$
V: $0.9^2 + 0.1^2 = 0.82$

In each case, the number of possible sets of answers to the first 3 questions is 8, of which 5 have $Y\not=1$ and therefore require a 4th question. I haven’t yet worked out the probability of going to heaven if you use my suggested strategy, but I assert that it is greater than 1/3.

We can start the calculation as follows.

Case 1
YNN has probability $0.625*0.58*0.82= 0.29725$.
YYN has probability $0.625*0.42*0.82= 0.21525$.
The probability that we choose the 2nd guard to put our 4th question to is 1/2, and then the probability of his saying N is 0.58 (heaven) and Y (0.42) (half chance of heaven); and the probability of our choosing the 1st guard is also 1/2, upon which the probability of his saying Y is 0.75 (half chance of heaven) and N (hell) is 0.25.
So given YYN our chance of getting to heaven is $0.21525 * \big(0.5*(0.58 +0.5*0.42)+(0.5*0.75)\big)=0.125383125$.
So already, having considered only YYN and YNN, we know that our probability of reaching heaven in Case 1 (door to heaven is guarded by Michael) is at least $0.21525+0.125383125=0.340633125$.
Since this is greater than 1/3, we have proved that in Case 1 the strategy is better than picking a door completely at random and I believe that that is also so in each of Cases 2 and 3 🙂