If` f_(n)(x),g_(n)(x),h_(n)(x),n=1, 2, 3` are polynomials in x such that `f_(n)(a)=g_(n)(a)=h_(n)(a),n=1,2,3` and
`F(x)=|{:(f_(1)(x),f_(2)(x),f_(3)(x)),(g_(1)(x),g_(2)(x),g_(3)(x)),(h_(1)(x),h_(2)(x),h_(3)(x)):}|`. Then, F' (a) is equal to

Options

A. 0
B. `f_(1)(a)g_(2)(a)h_(3)(a)`
C. 1
D. none of these

Explanation

To solve the problem step by step, we need to find the derivative \( F'(a) \) of the determinant defined by the polynomials \( f_n(x), g_n(x), h_n(x) \) at the point \( x = a \).

### Step 1: Define the Function
The function \( F(x) \) is given as the determinant of the matrix formed by the polynomials:
\[ F(x) = \begin{vmatrix} f_1(x) & f_2(x) & f_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1(x) & h_2(x) & h_3(x) \end{vmatrix} \]

### Step 2: Differentiate the Determinant
To differentiate \( F(x) \), we can use the property of determinants that allows us to differentiate one row at a time. The derivative \( F'(x) \) can be expressed as:
\[ F'(x) = \begin{vmatrix} f_1'(x) & f_2'(x) & f_3'(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1(x) & h_2(x) & h_3(x) \end{vmatrix} + \begin{vmatrix} f_1(x) & f_2(x) & f_3(x) \\ g_1'(x) & g_2'(x) & g_3'(x) \\ h_1(x) & h_2(x) & h_3(x) \end{vmatrix} + \begin{vmatrix} f_1(x) & f_2(x) & f_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1'(x) & h_2'(x) & h_3'(x) \end{vmatrix} \]

### Step 3: Evaluate at \( x = a \)
Now, we substitute \( x = a \) into the expression for \( F'(x) \):
\[ F'(a) = \begin{vmatrix} f_1'(a) & f_2'(a) & f_3'(a) \\ g_1(a) & g_2(a) & g_3(a) \\ h_1(a) & h_2(a) & h_3(a) \end{vmatrix} + \begin{vmatrix} f_1(a) & f_2(a) & f_3(a) \\ g_1'(a) & g_2'(a) & g_3'(a) \\ h_1(a) & h_2(a) & h_3(a) \end{vmatrix} + \begin{vmatrix} f_1(a) & f_2(a) & f_3(a) \\ g_1(a) & g_2(a) & g_3(a) \\ h_1'(a) & h_2'(a) & h_3'(a) \end{vmatrix} \]

### Step 4: Use the Given Conditions
Since we know that \( f_n(a) = g_n(a) = h_n(a) \) for \( n = 1, 2, 3 \), we can simplify the determinants:
- In each determinant, the rows will have identical values for \( f_n(a), g_n(a), h_n(a) \), leading to the determinant being zero.

Thus, we conclude:
\[ F'(a) = 0 + 0 + 0 = 0 \]

### Final Answer
\[ F'(a) = 0 \]