If the system of equations `(.^(n)C_(3))x+(.^(n)C_(4))y+35z=0, (.^(n)C_(4))x+35y+(.^(n)C_(3))z=0 and 35x+(.^(n)C_(3))y+(.^(n)C_(4))z=0` has a non - trivial solution, then the value of n is equal to `(AA n in N, n ge 4)`

Options

A. 6
B. 7
C. 8
D. 9

Answer

To solve the given system of equations for a non-trivial solution, we need to find the value of \( n \) such that the determinant of the coefficients matrix is zero. The equations are:

1. \( \binom{n}{3}x + \binom{n}{4}y + 35z = 0 \)
2. \( \binom{n}{4}x + 35y + \binom{n}{3}z = 0 \)
3. \( 35x + \binom{n}{3}y + \binom{n}{4}z = 0 \)

### Step 1: Form the Coefficient Matrix
The coefficient matrix \( A \) of the system is:

\[ A = \begin{bmatrix} \binom{n}{3} & \binom{n}{4} & 35 \\ \binom{n}{4} & 35 & \binom{n}{3} \\ 35 & \binom{n}{3} & \binom{n}{4} \end{bmatrix} \]

### Step 2: Calculate the Determinant
To find the value of \( n \) for which the system has a non-trivial solution, we need to compute the determinant of matrix \( A \) and set it to zero.

\[ \text{det}(A) = 0 \]

### Step 3: Use the Property of Combinations
For a non-trivial solution, we can use the property of combinations. We know that:

\[ \binom{n}{3} = \binom{n}{4} \text{ when } n = 3 + 4 = 7 \]

### Step 4: Set Conditions
We also need to ensure that both \( \binom{n}{3} \) and \( \binom{n}{4} \) equal 35:

\[ \binom{n}{3} = 35 \quad \text{and} \quad \binom{n}{4} = 35 \]

### Step 5: Solve for \( n \)
Using the formula for combinations:

1. \( \binom{n}{3} = \frac{n(n-1)(n-2)}{6} = 35 \)
- This simplifies to \( n(n-1)(n-2) = 210 \).

2. \( \binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{24} = 35 \)
- This simplifies to \( n(n-1)(n-2)(n-3) = 840 \).

### Step 6: Check Values of \( n \)
We can check \( n = 7 \):

- For \( n = 7 \):
- \( \binom{7}{3} = \frac{7 \cdot 6 \cdot 5}{6} = 35 \)
- \( \binom{7}{4} = \frac{7 \cdot 6 \cdot 5}{24} = 35 \)

Both conditions are satisfied.

### Conclusion
Thus, the value of \( n \) for which the system has a non-trivial solution is:

\[ \boxed{7} \]