The equation of a curve is $$ y=(2 k-3) x^{2}-k x-(k-2) $$, where $$ k $$ is a constant. The line $$ y=3 x-4 $$ is a tangent to the curve. Find the value of $$ k $$.

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1. Substitute the line equation $$ y=3x-4 $$ into the curve equation 
   $$
   y=(2k-3)x^2-kx-(k-2)
   $$
   to obtain:
   $$
   (2k-3)x^2 - kx - (k-2) = 3x-4.
   $$

2. Rearranging all terms to one side gives:
   $$
   (2k-3)x^2 - kx - (k-2) - 3x + 4 = 0.
   $$
   Combine like terms:
   $$
   (2k-3)x^2 - (k+3)x + [4-(k-2)] = 0.
   $$
   Simplify the constant term:
   $$
   4-(k-2)= -k+6.
   $$
   Hence, the quadratic in $$ x $$ is:
   $$
   (2k-3)x^2 - (k+3)x - k + 6 = 0.
   $$

3. For the line to be tangent to the curve, the quadratic must have exactly one solution, meaning its discriminant is zero. The discriminant $$\Delta$$ of a quadratic $$ ax^2+bx+c=0 $$ is given by:
   $$
   \Delta = b^2-4ac.
   $$
   Here, 
   $$
   a = 2k-3,\quad b = -(k+3),\quad c = -k+6.
   $$

4. Compute the discriminant:
   $$
   \Delta = (-(k+3))^2 - 4(2k-3)(-k+6).
   $$
   Simplify the expression:
   $$
   \Delta = (k+3)^2 + 4(2k-3)(k-6).
   $$

5. Expand $$(k+3)^2$$:
   $$
   (k+3)^2 = k^2+6k+9.
   $$

6. Expand $$ 4(2k-3)(k-6) $$:
   $$
   (2k-3)(k-6)=2k^2-12k-3k+18=2k^2-15k+18,
   $$
   so that
   $$
   4(2k-3)(k-6)=8k^2-60k+72.
   $$

7. Therefore, the discriminant becomes:
   $$
   \Delta = k^2+6k+9+8k^2-60k+72.
   $$
   Combine like terms:
   $$
   \Delta = 9k^2-54k+81.
   $$
   Factor out $$9$$:
   $$
   \Delta = 9(k^2-6k+9)=9(k-3)^2.
   $$

8. Set the discriminant equal to zero for tangency:
   $$
   9(k-3)^2=0.
   $$
   This implies:
   $$
   (k-3)^2=0.
   $$
   Thus,
   $$
   k-3=0 \quad \Longrightarrow \quad k=3.
   $$

The value of $$ k $$ is $$ \boxed{3} $$.
The value of $$ k $$ is 3.

The equation of a curve is , where is a constant. The line is a
tangent to the curve.
Find the value of .

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