Let $$\alpha$$ be the area of the larger region bounded by the curve $$y^2 = 8x$$ and the lines $$y = x$$ and $$x = 2$$, which lies in the first quadrant. Then the value of $$3\alpha$$ is equal to

Step 1: Find the points of intersection of the curves

The given curve is a parabola and the lines are:

$$y^2 = 8x$$

$$y = x$$

$$x = 2$$

Since the region lies in the first quadrant, we take the positive root of the parabola:

$$y = \sqrt{8x} = 2\sqrt{2}\sqrt{x}$$

To find where the parabola $$y^2 = 8x$$ and the line $$y = x$$ intersect, we equate them:

$$x^2 = 8x$$

$$x(x - 8) = 0$$

$$x = 0 \quad \text{or} \quad x = 8$$

Thus, the parabola and the line intersect at the points $$(0, 0)$$ and $$(8, 8)$$.

Step 2: Analyze the regions formed by the boundary line $$x = 2$$

The vertical line $$x = 2$$ divides the area enclosed between the curve $$y = 2\sqrt{2}\sqrt{x}$$ and the line $$y = x$$ into two separate regions within the first quadrant:

- Region 1 (Left region): From $$x = 0$$ to $$x = 2$$

- Region 2 (Right region): From $$x = 2$$ to $$x = 8$$

Step 3: Calculate the area of both regions to identify the larger one

For Region 1, the area $$A_1$$ is bounded from $$x = 0$$ to $$x = 2$$:

$$A_1 = \int_0^2 (2\sqrt{2}\sqrt{x} - x) \, dx$$

$$A_1 = \left[ 2\sqrt{2} \cdot \frac{2}{3}x^{3/2} - \frac{x^2}{2} \right]_0^2$$

$$A_1 = \left[ \frac{4\sqrt{2}}{3}x\sqrt{x} - \frac{x^2}{2} \right]_0^2$$

$$A_1 = \left( \frac{4\sqrt{2}}{3}(2\sqrt{2}) - \frac{4}{2} \right) - 0$$

$$A_1 = \frac{16}{3} - 2 = \frac{10}{3}$$

For Region 2, the area $$A_2$$ is bounded from $$x = 2$$ to $$x = 8$$:

$$A_2 = \int_2^8 (2\sqrt{2}\sqrt{x} - x) \, dx$$

$$A_2 = \left[ \frac{4\sqrt{2}}{3}x\sqrt{x} - \frac{x^2}{2} \right]_2^8$$

$$A_2 = \left( \frac{4\sqrt{2}}{3}(8\sqrt{8}) - \frac{64}{2} \right) - \left( \frac{16}{3} - 2 \right)$$

$$A_2 = \left( \frac{4\sqrt{2}}{3}(16\sqrt{2}) - 32 \right) - \frac{10}{3}$$

$$A_2 = \left( \frac{128}{3} - 32 \right) - \frac{10}{3}$$

$$A_2 = \frac{118}{3} - 32$$

$$A_2 = \frac{118 - 96}{3} = \frac{22}{3}$$

Step 4: Identify the larger region $$\alpha$$ and calculate $$3\alpha$$

Comparing the two areas:

$$\alpha = \max(A_1, A_2) = \max\left(\frac{10}{3}, \frac{22}{3}\right) = \frac{22}{3}$$

Now, compute the required value:

$$3\alpha = 3 \cdot \frac{22}{3} = 22$$

Conclusion:

The value of $$3\alpha$$ is equal to 22.