Question:

Step 1: Evaluate the given determinant. The given determinant is: $$ \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ \alpha & \beta & \gamma \end{vmatrix} = t $$ We expand the determinant along the first row: $$ 1(2\gamma - 3\beta) - 1(\gamma - 3\alpha) + 1(\beta - 2\alpha) = t $$ $$ 2\gamma - 3\beta - \gamma + 3\alpha + \beta - 2\alpha = t $$ $$ \alpha - 2\beta + \gamma = t $$ Step 2: Determine the value of $t$. The problem states that $t$ is an even prime number. The only even prime number is $2$. Therefore, we set $t=2$. So, the equation from the determinant evaluation becomes: $$ \alpha - 2\beta + \gamma = 2 $$ Step 3: Apply Vieta's formulas to the given cubic equation. The given cubic equation is $x^3 - 14x^2 + Px - 36 = 0$, and its integral roots are $\alpha, \beta, \gamma$. According to Vieta's formulas for a cubic equation $ax^3 + bx^2 + cx + d = 0$ with roots $r_1, r_2, r_3$: Sum of roots: $r_1 + r_2 + r_3 = -b/a$ Sum of products of roots taken two at a time: $r_1r_2 + r_2r_3 + r_3r_1 = c/a$ Product of roots: $r_1r_2r_3 = -d/a$ Applying these to the given equation: $$ \alpha + \beta + \gamma = -(-14)/1 = 14 \quad \text{(Equation 1)} $$ $$ \alpha\beta\gamma = -(-36)/1 = 36 \quad \text{(Equation 2)} $$ $$ \alpha\beta + \beta\gamma + \gamma\alpha = P/1 = P \quad \text{(Equation 3)} $$ Step 4: Solve for $\beta$ using the system of equations. We have the equation derived from the determinant: $$ \alpha - 2\beta + \gamma = 2 \quad \text{(Equation 4)} $$ And from Vieta's formulas: $$ \alpha + \beta + \gamma = 14 \quad \text{(Equation 1)} $$ Subtract Equation 4 from Equation 1: $$ (\alpha + \beta + \gamma) - (\alpha - 2\beta + \gamma) = 14 - 2 $$ $$ \alpha + \beta + \gamma - \alpha + 2\beta - \gamma = 12 $$ $$ 3\beta = 12 $$ $$ \beta = 4 $$ Step 5: Find the remaining roots $\alpha$ and $\gamma$. Substitute $\beta = 4$ into Equation 1: $$ \alpha + 4 + \gamma = 14 $$ $$ \alpha + \gamma = 10 $$ Now substitute $\beta = 4$ into Equation 2: $$ \alpha(4)\gamma = 36 $$ $$ \alpha\gamma = 9 $$ We now have two equations: $\alpha + \gamma = 10$ and $\alpha\gamma = 9$. This means $\alpha$ and $\gamma$ are the roots of the quadratic equation $y^2 - (\alpha+\gamma)y + \alpha\gamma = 0$: $$ y^2 - 10y + 9 = 0 $$ Factoring the quadratic equation: $$ (y-1)(y-9) = 0 $$ The roots are $y=1$ and $y=9$. Therefore, the three integral roots of the cubic equation are $1, 4, 9$. We can assign $\alpha=1, \beta=4, \gamma=9$ (or any permutation). Step 6: Calculate the value of $P$. From Equation 3, $P = \alpha\beta + \beta\gamma + \gamma\alpha$. Substitute the values of the roots $1, 4, 9$: $$ P = (1)(4) + (4)(9) + (9)(1) $$ $$ P = 4 + 36 + 9 $$ $$ P = 49 $$ Step 7: Classify the value of $P$ and select the correct option. The calculated value of $P$ is $49$. Let's check the given options: (A) a rational number: $49$ is a rational number. (B) a prime number: $49 = 7^2$, so it is not a prime number. (C) an odd natural number: $49$ is a positive integer not divisible by $2$, making it an odd natural number. (D) an even natural number: $49$ is not an even natural number. While $P=49$ is a rational number, option (C) provides a more specific classification. The final answer is $\boxed{\text{an odd natural number}}$.