The equations of the sides AB, BC and CA of a triangle ABC are $$2x + y = 0$$, $$x + py = 39$$ and $$x - y = 3$$ respectively and P(2,3) is its circumcentre. Then which of the following is NOT true

Vertex $$A$$ is the intersection of $$AB: 2x+y=0$$ and $$CA: x-y=3$$. Adding: $$3x = 3 \implies x = 1, y = -2$$, so $$A = (1, -2)$$.

Vertex $$C$$ is the intersection of $$BC: x+py=39$$ and $$CA: x-y=3$$. From $$x = y+3$$: $$(y+3)+py = 39 \implies y(1+p) = 36 \implies y = \dfrac{36}{1+p}$$, giving $$C = \left(\dfrac{39+3p}{1+p},\ \dfrac{36}{1+p}\right)$$.

Vertex $$B$$ is the intersection of $$AB: 2x+y=0$$ and $$BC: x+py=39$$. From $$y = -2x$$: $$x - 2px = 39 \implies x = \dfrac{39}{1-2p}$$, hence $$B = \left(\dfrac{39}{1-2p},\ \dfrac{-78}{1-2p}\right)$$.

The circumcentre $$P(2,3)$$ is equidistant from all vertices, and

$$PA^2 = (2-1)^2 + (3+2)^2 = 1 + 25 = 26$$.

Setting $$PA^2 = PC^2$$:

$$26 = \left(2 - \frac{39+3p}{1+p}\right)^2 + \left(3 - \frac{36}{1+p}\right)^2 = \frac{(-37-p)^2 + (3p-33)^2}{(1+p)^2}$$

Expanding:

$$26(1+p)^2 = (37+p)^2 + (33-3p)^2$$ $$26 + 52p + 26p^2 = 1369 + 74p + p^2 + 1089 - 198p + 9p^2$$ $$16p^2 + 176p - 2432 = 0 \implies p^2 + 11p - 152 = 0$$ $$(p - 8)(p + 19) = 0$$

$$p = -19$$ makes $$B = A$$ (degenerate), so $$p = 8$$.

With $$p = 8$$: $$B = \left(-\dfrac{13}{5}, \dfrac{26}{5}\right)$$, $$C = (7, 4)$$.

$$PB^2 = \left(2+\dfrac{13}{5}\right)^2 + \left(3-\dfrac{26}{5}\right)^2 = \left(\dfrac{23}{5}\right)^2 + \left(-\dfrac{11}{5}\right)^2 = \dfrac{529+121}{25} = 26 = PA^2$$ $$\checkmark$$

$$AC^2 = (7-1)^2 + (4+2)^2 = 36 + 36 = 72$$.

Option A: $$AC^2 = 72 = 9 \times 8 = 9p$$ $$\checkmark$$

Option B: $$AC^2 + p^2 = 72 + 64 = 136$$ $$\checkmark$$

Area of $$\triangle ABC$$:

$$\text{Area} = \frac{1}{2}\left|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)\right|$$ $$= \frac{1}{2}\left|1\left(\frac{26}{5} - 4\right) + \left(-\frac{13}{5}\right)(4+2) + 7\left(-2-\frac{26}{5}\right)\right|$$ $$= \frac{1}{2}\left|\frac{6}{5} - \frac{78}{5} - \frac{252}{5}\right| = \frac{1}{2} \times \frac{324}{5} = \frac{162}{5} = 32.4$$

Option C: $$32 < 32.4 < 36$$ $$\checkmark$$

Option D: $$34 < 32.4 < 38$$ $$\boldsymbol{\times}$$

The statement that is NOT true is $$\boxed{34 < \text{area}(\triangle ABC) < 38}$$. The answer is Option D.