Number of positive integral solutions of xyz = 30 is (a) 9(b) 27(c) 81(d) 243

To solve the problem of finding the number of positive integral solutions to the equation \(xyz = 30\), follow these steps:

### Step 1: Prime Factorization of 30
Prime factorize 30:
30 = \(2^1 \times 3^1 \times 5^1\)

Each variable \(x\), \(y\), and \(z\) will share the powers of these prime factors in some way.

### Step 2: Distribute Powers
For each prime factor, the sum of the powers in \(x\), \(y\), and \(z\) must equal the total power of that factor in 30. Let us consider the distribution for each prime factor.

#### For prime factor 2:
The power of 2 in \(x\), \(y\), and \(z\) must add up to 1:
\[a_1 + b_1 + c_1 = 1\]

The number of non-negative integral solutions of this equation is given by the stars-and-bars formula:
\[
\binom{1 + 3 - 1}{3 - 1} = \binom{3}{2} = 3
\]

#### For prime factor 3:
The power of 3 in \(x\), \(y\), and \(z\) must add up to 1:
\[a_2 + b_2 + c_2 = 1\]

Similarly, the number of solutions is:
\[
\binom{1 + 3 - 1}{3 - 1} = \binom{3}{2} = 3
\]

#### For prime factor 5:
The power of 5 in \(x\), \(y\), and \(z\) must add up to 1:
\[a_3 + b_3 + c_3 = 1\]

The number of solutions is:
\[
\binom{1 + 3 - 1}{3 - 1} = \binom{3}{2} = 3
\]

### Step 3: Combine Results
The total number of positive integral solutions is the product of the solutions for each prime factor, because the choices for distributing the powers of 2, 3, and 5 are independent:
\[
3 \times 3 \times 3 = 27
\]

### Final Answer:
The number of positive integral solutions of \(xyz = 30\) is **27**.

Correct option: (b) 27