Key Concepts and Formulas

• Vertices of a Triangle: The vertices are the intersection points of the sides.
• Centroid of a Triangle: If the vertices are $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$, the centroid $G(X, Y)$ is given by $X = \frac{x_1+x_2+x_3}{3}$ and $Y = \frac{y_1+y_2+y_3}{3}$.
• Distance Formula: The square of the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2$.

Step 1: Determine the Coordinates of Vertex A

Vertex A is the intersection of sides AB and CA.

• Side AB: $2x+y=0$
• Side CA: $x-y=3$

Why this step? We need the coordinates of all three vertices to use the centroid formula. Finding A first is convenient because its defining equations don't involve the parameters 'a' and 'p'.

Solving the system of equations: From the equation of AB, we have $y = -2x$. Substituting this into the equation of CA, we get $x - (-2x) = 3$, which simplifies to $3x = 3$, so $x = 1$. Substituting $x=1$ back into $y=-2x$, we get $y = -2(1) = -2$. Therefore, the coordinates of vertex A are $(1, -2)$.

Step 2: Express the Coordinates of Vertices B and C in terms of 'a'

Let the coordinates of B be $(x_B, y_B)$ and C be $(x_C, y_C)$. B lies on AB: $2x_B + y_B = 0 \implies y_B = -2x_B$ C lies on CA: $x_C - y_C = 3 \implies y_C = x_C - 3$ The centroid P is given as $(2, a)$.

Why this step? Expressing B and C in terms of 'a' allows us to utilize the centroid information effectively. This avoids the complexity of directly solving for B and C in terms of both 'a' and 'p' initially.

Using the centroid formula: $2 = \frac{1 + x_B + x_C}{3} \implies x_B + x_C = 5$ (1) $a = \frac{-2 + y_B + y_C}{3} \implies y_B + y_C = 3a + 2$ (2)

Substituting $y_B = -2x_B$ and $y_C = x_C - 3$ into equation (2): $-2x_B + x_C - 3 = 3a + 2 \implies -2x_B + x_C = 3a + 5$ (3)

Now we have a system of two equations: (1) $x_B + x_C = 5$ (3) $-2x_B + x_C = 3a + 5$

Subtracting equation (3) from equation (1): $(x_B + x_C) - (-2x_B + x_C) = 5 - (3a + 5) \implies 3x_B = -3a \implies x_B = -a$ Substituting $x_B = -a$ into equation (1): $-a + x_C = 5 \implies x_C = 5 + a$

Now find $y_B$ and $y_C$: $y_B = -2x_B = -2(-a) = 2a$ $y_C = x_C - 3 = (5 + a) - 3 = a + 2$

So, $B(-a, 2a)$ and $C(5+a, a+2)$.

Step 3: Determine the values of 'a' and 'p'

Both B and C lie on the line BC: $x + py = 21a$.

Why this step? By substituting the coordinates of B and C into the equation of BC, we create a system of equations that allows us to solve for the unknown parameters 'a' and 'p'.

Substituting $B(-a, 2a)$ into $x + py = 21a$: $-a + p(2a) = 21a$. Since $a \neq 0$, we can divide by $a$: $-1 + 2p = 21 \implies 2p = 22 \implies p = 11$

Substituting $C(5+a, a+2)$ into $x + py = 21a$: $(5+a) + p(a+2) = 21a$. Substituting $p = 11$: $5 + a + 11(a+2) = 21a \implies 5 + a + 11a + 22 = 21a \implies 12a + 27 = 21a \implies 9a = 27 \implies a = 3$

So, $a = 3$ and $p = 11$.

Step 4: Calculate the specific coordinates of Vertices B and C

Now that we have $a=3$, we can find the numerical coordinates of B and C.

Why this step? We need the specific numerical coordinates of B and C to calculate the distance between them.

Using $B(-a, 2a)$: $B(-3, 2(3)) \implies B(-3, 6)$

Using $C(5+a, a+2)$: $C(5+3, 3+2) \implies C(8, 5)$

Step 5: Determine if a = 3 is correct We have A(1, -2), B(-3, 6), C(8, 5). Check that the centroid is P(2, 3).

Why this step? Double-checking the value of 'a' is crucial.

$X_P = \frac{1 - 3 + 8}{3} = \frac{6}{3} = 2$ $Y_P = \frac{-2 + 6 + 5}{3} = \frac{9}{3} = 3$ P(2, 3) is the centroid, so a = 3 is correct.

Step 6: Calculate $(BC)^2$

Using the distance formula for $B(-3, 6)$ and $C(8, 5)$:

Why this step? This is the final calculation required by the problem statement.

$(BC)^2 = (8 - (-3))^2 + (5 - 6)^2 = (11)^2 + (-1)^2 = 121 + 1 = 122$

Step 7: Verify the value of p Since C(8, 5) lies on BC: x + py = 21a, we have 8 + 11(5) = 21(3) => 8 + 55 = 63 => 63 = 63. Hence, the value of p is correct.

Common Mistakes & Tips

• Always double-check algebraic manipulations to avoid sign errors or miscalculations.
• Remember to use the given information effectively. The condition $a \neq 0$ is crucial for simplifying equations.
• After finding the values of parameters, substitute them back to ensure consistency and correctness.

Summary

We found the vertices A, B, and C by intersecting the given lines. Using the centroid formula, we expressed the coordinates of B and C in terms of 'a'. Then, using the equation of line BC, we solved for 'a' and 'p'. Finally, we calculated the square of the distance BC.

The final answer is \boxed{122}. However, the options listed in the prompt do not include 122. Therefore, there may be an error in the problem statement, options, or the given correct answer.

Given that the "Correct Answer" is stated as 21, let's work backwards and see if we can arrive at that solution.

If $(BC)^2 = 21$, then $BC = \sqrt{21}$. So, $(x_C - x_B)^2 + (y_C - y_B)^2 = 21$. Substituting $x_B = -a$, $y_B = 2a$, $x_C = 5+a$, $y_C = a+2$:

$(5+a+a)^2 + (a+2-2a)^2 = 21$ $(5+2a)^2 + (2-a)^2 = 21$ $25 + 20a + 4a^2 + 4 - 4a + a^2 = 21$ $5a^2 + 16a + 29 = 21$ $5a^2 + 16a + 8 = 0$

$a = \frac{-16 \pm \sqrt{256 - 160}}{10} = \frac{-16 \pm \sqrt{96}}{10} = \frac{-16 \pm 4\sqrt{6}}{10} = \frac{-8 \pm 2\sqrt{6}}{5}$

This leads to a non-integer value for 'a', which seems less likely given the problem context. Thus, it's most probable that there's an error in the provided correct answer.

The problem asks for the square of the distance and the original answer of 122 seems much more plausible.

Given the discrepancy between the derived answer and the provided "Correct Answer", I will maintain the derived answer of 122 and report the discrepancy.

The final answer is \boxed{122}.