Let $$l$$ be a line which is normal to the curve $$y = 2x^2 + x + 2$$ at a point $$P$$ on the curve. If the point $$Q(6, 4)$$ lies on the line $$l$$ and $$O$$ is origin, then the area of the triangle $$OPQ$$ is equal to ______

The curve is $$y = 2x^2 + x + 2$$. At point $$P(a, 2a^2 + a + 2)$$, the slope of the tangent is:

$$\frac{dy}{dx} = 4x + 1 \implies \text{slope at } P = 4a + 1$$

The normal at $$P$$ has slope $$= -\frac{1}{4a+1}$$.

Since $$Q(6, 4)$$ lies on the normal:

$$\frac{4 - (2a^2 + a + 2)}{6 - a} = -\frac{1}{4a+1}$$

$$\frac{2 - 2a^2 - a}{6 - a} = -\frac{1}{4a+1}$$

Cross-multiplying:

$$(2 - 2a^2 - a)(4a + 1) = -(6 - a)$$

$$(2 - 2a^2 - a)(4a+1) = a - 6$$

Expanding the left side:

$$8a + 2 - 8a^3 - 2a^2 - 4a^2 - a = a - 6$$

$$-8a^3 - 6a^2 + 7a + 2 = a - 6$$

$$-8a^3 - 6a^2 + 6a + 8 = 0$$

$$8a^3 + 6a^2 - 6a - 8 = 0$$

$$4a^3 + 3a^2 - 3a - 4 = 0$$

Testing $$a = 1$$: $$4 + 3 - 3 - 4 = 0$$. So $$a = 1$$ is a root.

Factoring: $$4a^3 + 3a^2 - 3a - 4 = (a - 1)(4a^2 + 7a + 4)$$

The discriminant of $$4a^2 + 7a + 4$$ is $$49 - 64 = -15 < 0$$, so no other real roots.

Thus $$P = (1, 5)$$.

Area of triangle $$OPQ$$ with $$O(0,0)$$, $$P(1, 5)$$, $$Q(6, 4)$$:

$$\text{Area} = \frac{1}{2}|x_O(y_P - y_Q) + x_P(y_Q - y_O) + x_Q(y_O - y_P)|$$

$$= \frac{1}{2}|0(5-4) + 1(4-0) + 6(0-5)|$$

$$= \frac{1}{2}|0 + 4 - 30| = \frac{1}{2} \times 26 = 13$$

Hence the answer is $$\boxed{13}$$.