The area of the region enclosed between the parabola $y = \frac{3x^2}{4}$ and the line $3x - 2y = 12$ is,

The region enclosed between the parabola $y = \frac{3x^2}{4}$ and the line $3x - 2y = 12$ requires finding their intersection points.

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The line equation $3x - 2y = 12$ can be rewritten in slope-intercept form:

$3x - 2y = 12$

$-2y = 12 - 3x$

$y = \frac{3x - 12}{2}$

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For intersection points, set the two expressions for $y$ equal:

$\frac{3x^2}{4} = \frac{3x - 12}{2}$

Multiplying both sides by 4:

$3x^2 = 2(3x - 12)$

$3x^2 = 6x - 24$

$3x^2 - 6x + 24 = 0$

Dividing by 3:

$x^2 - 2x + 8 = 0$

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For the quadratic equation $x^2 - 2x + 8 = 0$, the discriminant is:

$\Delta = b^2 - 4ac$

$\Delta = (-2)^2 - 4(1)(8)$

$\Delta = 4 - 32$

$\Delta = -28$

Since $\Delta < 0$, there are no real solutions. The parabola and line do not intersect.

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Without intersection points, the curves do not enclose a region. Therefore, no area can be calculated, and the question is mathematically invalid.

Welcome to NTA's world, we're just living in it. The question was dropped and everyone was awarded 5 marks.