Consider a curve $$y = y(x)$$ in the first quadrant as shown in the figure. Let the area $$A_1$$ is twice the area $$A_2$$. Then the normal to the curve perpendicular to the line $$2x - 12y = 15$$ does NOT pass through the point

We are given a curve $$y = y(x)$$ in the first quadrant, with area $$A_1$$ being twice area $$A_2$$. We need to find which point the normal (perpendicular to $$2x - 12y = 15$$) does NOT pass through.

From the figure description, for a point $$(x, y)$$ on the curve in the first quadrant:

$$A_1$$ = area under the curve from 0 to $$x$$ = $$\int_0^x y\,dx$$

$$A_2$$ = area to the left of the curve from 0 to $$y$$ = $$\int_0^y x\,dy$$

Since $$A_1 = 2A_2$$, and we know $$A_1 + A_2 = xy$$ (the rectangle), we get $$2A_2 + A_2 = xy$$, so $$A_2 = \frac{xy}{3}$$.

Therefore $$A_1 = \frac{2xy}{3}$$.

From $$A_1 = \int_0^x y\,dx = \frac{2xy}{3}$$.

Differentiating both sides with respect to $$x$$:

Since the curve passes through the first quadrant and using the standard form, $$y = k\sqrt{x}$$.

The line $$2x - 12y = 15$$ has slope $$\frac{2}{12} = \frac{1}{6}$$.

The normal is perpendicular to this line, so the normal has slope $$-6$$.

Since $$y = k\sqrt{x}$$, $$y' = \frac{k}{2\sqrt{x}}$$.

The slope of the normal at any point is $$-\frac{1}{y'} = -\frac{2\sqrt{x}}{k}$$.

Setting this equal to $$-6$$:

At this point, $$y = k\sqrt{9k^2} = k \cdot 3k = 3k^2$$.

The normal passes through $$(9k^2, 3k^2)$$ with slope $$-6$$:

For the line $$y = -6x + 57k^2$$:

Option A: $$(6, 21)$$: $$21 = -36 + 57k^2 \Rightarrow 57k^2 = 57 \Rightarrow k^2 = 1$$. Valid.

Option B: $$(8, 9)$$: $$9 = -48 + 57k^2 \Rightarrow 57k^2 = 57 \Rightarrow k^2 = 1$$. Valid.

Option C: $$(10, -4)$$: $$-4 = -60 + 57k^2 \Rightarrow 57k^2 = 56 \Rightarrow k^2 = \frac{56}{57}$$. NOT a clean value.

Option D: $$(12, -15)$$: $$-15 = -72 + 57k^2 \Rightarrow 57k^2 = 57 \Rightarrow k^2 = 1$$. Valid.

Options A, B, and D all give $$k^2 = 1$$, meaning the normal passes through them. Option C does NOT lie on any such normal.

The correct answer is Option C: $$(10, -4)$$.