The equation of a curve is $y = ax^2 - 3x + 4 - a$, where $a$ is a constant.

1. Substitute $a = -2$ into the equation of the curve and set up the inequality

When $a = -2$, the equation of the curve is:
$y = -2x^2 - 3x + 4 - (-2)$
$y = -2x^2 - 3x + 6$

For the curve to lie completely below the line $y = -3$:
$-2x^2 - 3x + 6 < -3$

2. Rearrange the inequality into standard quadratic form

$-2x^2 - 3x + 9 < 0$
Multiply by $-1$ and flip the inequality sign:
$2x^2 + 3x - 9 > 0$

3. Factorise and solve the quadratic inequality

Factorising the quadratic expression:
$(2x - 3)(x + 3) > 0$

The critical values are $x = 1.5$ and $x = -3$.
Since the quadratic is positive ($> 0$), the solution is:
$x < -3 \text{ or } x > 1.5$

1. Substitute $a = 3$ into the curve equation and equate it to the line equation

When $a = 3$, the equation of the curve is:
$y = 3x^2 - 3x + 4 - 3$
$y = 3x^2 - 3x + 1$

Equating the line and the curve:
$3x^2 - 3x + 1 = 3x - 2$

2. Rearrange into a standard quadratic equation and evaluate its discriminant

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

Discriminant, $b^2 - 4ac = (-6)^2 - 4(3)(3)$
$= 36 - 36$
$= 0$

Since the discriminant is $0$, the equation has real and equal roots. Thus, the line intersects the curve at exactly one point, which means it is a tangent to the curve.

1. Form the equation for the intersection of the curve and the line.

$ax^2 - 3x + 4 - a = 3x - 2$
$ax^2 - 6x + 6 - a = 0$

2. Apply the condition for the line to intersect the curve at exactly one point (tangent).

For one point of intersection, the discriminant must be zero:
$b^2 - 4ac = 0$
$(-6)^2 - 4(a)(6 - a) = 0$
$36 - 24a + 4a^2 = 0$
$4a^2 - 24a + 36 = 0$
$a^2 - 6a + 9 = 0$

3. Solve for $a$ and conclude.

$(a - 3)^2 = 0$
$a = 3$
Since $a = 3$ is the only solution, there is no other value of $a$.