If `Delta=|(1,1,1),(""^(m)C_(1),""^(m+3)C_(1),""^(m+6)C_(1)),(""^(m)C_(2),""^(m+3)C_(2),""^(m+6)C_(2))|=2^(alpha)3^(beta),5^(gamma)`, then `alpha+beta+gamma` is equal

Options

A. 3
B. 5
C. 7
D. None of these

Answer

To solve the given determinant problem step-by-step, we will evaluate the determinant \( \Delta \) defined as:

\[ \Delta = \begin{vmatrix} 1 & \binom{m}{1} & \binom{m+3}{1} \\ \binom{m}{2} & \binom{m+3}{2} & \binom{m+6}{2} \end{vmatrix} \]

### Step 1: Calculate the binomial coefficients

Using the formula for binomial coefficients, we have:

- \( \binom{m}{1} = m \)
- \( \binom{m+3}{1} = m + 3 \)
- \( \binom{m+6}{1} = m + 6 \)
- \( \binom{m}{2} = \frac{m(m-1)}{2} \)
- \( \binom{m+3}{2} = \frac{(m+3)(m+2)}{2} \)
- \( \binom{m+6}{2} = \frac{(m+6)(m+5)}{2} \)

Substituting these values into the determinant, we get:

\[ \Delta = \begin{vmatrix} 1 & m & m + 3 \\ \frac{m(m-1)}{2} & \frac{(m+3)(m+2)}{2} & \frac{(m+6)(m+5)}{2} \end{vmatrix} \]

### Step 2: Factor out common terms

We can factor out \( \frac{1}{2} \) from the second row:

\[ \Delta = \frac{1}{2} \begin{vmatrix} 1 & m & m + 3 \\ m(m-1) & (m+3)(m+2) & (m+6)(m+5) \end{vmatrix} \]

### Step 3: Simplify the determinant

Next, we will perform column operations to simplify the determinant. We can replace the second and third columns with the difference of those columns and the first column:

- Replace column 2 with column 2 - column 1
- Replace column 3 with column 3 - column 1

This gives us:

\[ \Delta = \frac{1}{2} \begin{vmatrix} 1 & 0 & 0 \\ m(m-1) & (m+3)(m+2) - m(m-1) & (m+6)(m+5) - m(m-1) \end{vmatrix} \]

Calculating the new entries in the second row:

1. For the second column:
\[ (m+3)(m+2) - m(m-1) = m^2 + 5m + 6 - (m^2 - m) = 6m + 6 = 6(m + 1) \]

2. For the third column:
\[ (m+6)(m+5) - m(m-1) = m^2 + 11m + 30 - (m^2 - m) = 12m + 30 \]

Thus, the determinant simplifies to:

\[ \Delta = \frac{1}{2} \begin{vmatrix} 1 & 0 & 0 \\ m(m-1) & 6(m + 1) & 12m + 30 \end{vmatrix} \]

### Step 4: Calculate the determinant

The determinant can now be calculated as:

\[ \Delta = \frac{1}{2} \cdot 1 \cdot \left( m(m-1) \cdot 12(m + 1) - 0 \right) = \frac{1}{2} \cdot 12m(m-1)(m+1) = 6m(m^2 - 1) \]

### Step 5: Factor and express in terms of powers of 2, 3, and 5

Now, we need to express \( \Delta \) in the form \( 2^\alpha 3^\beta 5^\gamma \):

\[ \Delta = 6m(m^2 - 1) = 6m(m-1)(m+1) = 2 \cdot 3 \cdot m(m-1)(m+1) \]

### Step 6: Identify \( \alpha, \beta, \gamma \)

From \( 6 = 2^1 \cdot 3^1 \):

- \( \alpha = 1 \) (from the factor of 2)
- \( \beta = 1 \) (from the factor of 3)
- \( \gamma = 0 \) (no factor of 5)

Thus, we have:

\[ \alpha + \beta + \gamma = 1 + 1 + 0 = 2 \]

### Final Answer

The value of \( \alpha + \beta + \gamma \) is:

\[ \boxed{2} \]