I will go with 6 i.e option B.

I course contain 3 varieties

II course contain 4 varieties

III course contain 3 varieties

Suppose there are p beverages .

We have to order a dish for atleat two courses and one beverage.

Case 1 Exactly 2 courses and one beverage

These 2 courses again could be from

I ,II or II, III or I, III

So this will be 3C1x4C1xpC1 + 4C1x3C1xpC1 + 3C1x3C1xpC1

i.e 12p+12p+9p = 33p

Case 2: Exactly 3 courses and one beverage.

This will be 3C1x4C1x3C1xpC1 = 36p

So 33p+36p=365

i.e 69p=365

So p>5

So minimum 6 beverages must be offered

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Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)