\begin{array}{l}\text{If}\mathrm{ 3}x+y=8\text{ is a tangent to the curve}{\mathrm{ y}}^2=\alpha +\beta x^3\text{ at} (2,2),\text{then value of}\mathrm{ \alpha }-\beta \text{ 12is}\end{array}

The correct option is 13.

We are given that the line
$$3x+y=8$$
is tangent to the curve
$$y^2=\alpha +\beta x^3$$
at the point (2, 2).

First, since the point (2, 2) lies on the curve, it must satisfy the equation of the curve. Substituting x = 2 and y = 2 into the curve equation:
$$2^2=\alpha +\beta {\left(2\right)}^3$$
$$4=\alpha +8\beta (Equation\; 1)$$

Next, we find the slope of the tangent to the curve at the point (2, 2). We do this by differentiating both sides of the curve's equation with respect to x:
$$\frac{d}{dx}\left(y^2\right)=\frac{d}{dx}(\mathrm{��}+\beta x^3)$$
$$2y\frac{dy}{dx}=3\beta x^2$$
$$\frac{dy}{dx}=\frac{3\beta x^2}{2y}$$

Substituting the coordinates of the point of tangency (2, 2) to find the slope of the tangent line:
$${\left(\frac{dy}{dx}\right)}_{(2,2)}=\frac{3\beta {\left(2\right)}^2}{2\left(2\right)}$$

We are also given the equation of the tangent line as:
$$3x+y=8\Rightarrow y=-3x+8$$
The slope of this line is -3.

Equating the slope obtained from differentiation to the slope of the given line:
$$3\beta =-3\Rightarrow \beta =-1$$

Now, substitute the value of β = -1 back into Equation 1 to find α:
$$4=\alpha +8(-1)$$
$$4=\alpha -8\Rightarrow \alpha =12$$

We need to find the value of $\alpha -\beta$ (noting that the typo in the question text "α - β 12is" refers to the value of α - β):
$$\alpha -\beta =12-(-1)=12+1=13$$