If `px^(4)+qx^(3)+rx^(2)+sx+t=|{:(x^(2)+3x,x-1,x+3),(x^(2)+1,2-x,x-3),(x^(2)-3,x+4,3x):}|` then t is equal to

Options

A. `-5`
B. `21`
C. `-3`
D. `-2`

Answer

To solve the problem, we need to evaluate the determinant given in the question and find the value of \( t \) when \( x = 0 \).

### Step-by-step Solution:

1. **Set Up the Determinant**:
The determinant we need to evaluate is:
\[ \begin{vmatrix} x^2 + 3x & x - 1 & x + 3 \\ x^2 + 1 & 2 - x & x - 3 \\ x^2 - 3 & x + 4 & 3x \end{vmatrix} \]

2. **Substitute \( x = 0 \)**:
To find \( t \), we substitute \( x = 0 \) into the determinant:
\[ \begin{vmatrix} 0^2 + 3(0) & 0 - 1 & 0 + 3 \\ 0^2 + 1 & 2 - 0 & 0 - 3 \\ 0^2 - 3 & 0 + 4 & 3(0) \end{vmatrix} = \begin{vmatrix} 0 & -1 & 3 \\ 1 & 2 & -3 \\ -3 & 4 & 0 \end{vmatrix} \]

3. **Calculate the Determinant**:
We can expand the determinant using the first row:
\[ = 0 \cdot \begin{vmatrix} 2 & -3 \\ 4 & 0 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & -3 \\ -3 & 0 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 2 \\ -3 & 4 \end{vmatrix} \]
The first term is zero because it is multiplied by zero.

Now, calculate the second term:
\[ = -(-1) \cdot (1 \cdot 0 - (-3) \cdot (-3)) = 1 \cdot (0 - 9) = -9 \]

Now, calculate the third term:
\[ = 3 \cdot (1 \cdot 4 - 2 \cdot (-3)) = 3 \cdot (4 + 6) = 3 \cdot 10 = 30 \]

Combine the results:
\[ = 0 + 9 + 30 = 39 \]

4. **Final Value of \( t \)**:
Therefore, the value of \( t \) is:
\[ t = 39 \]

### Conclusion:
The value of \( t \) is \( 39 \).