A rod of length eight units moves such that its ends A and B always lie on the lines x - y + 2=0 and y + 2 = 0. respectively. If the locus of the point P, that divides the rod AB internally in the ratio 2:1 is $$9(x^{2}+\alpha y^{2}+\beta xy+\gamma x+ 28y)-76=0$$. then $$\alpha -\beta -\gamma$$ equals to :

A rod of length 8 units moves with ends A on $$x - y + 2 = 0$$ and B on $$y + 2 = 0$$. Point P divides AB internally in ratio 2:1.

Line $$x - y + 2 = 0$$ can be written as $$y = x + 2$$, so we set $$A = (a, a+2)$$.

On the other hand, $$y + 2 = 0$$ gives $$y = -2$$, and we take $$B = (b, -2)$$.

By the section formula, the point P that divides AB in the ratio 2:1 is $$P = \left(\frac{2b + a}{3}, \frac{2(-2) + (a+2)}{3}\right) = \left(\frac{2b + a}{3}, \frac{a - 2}{3}\right)$$.

Writing $$x = \frac{2b + a}{3}$$ and $$y = \frac{a - 2}{3}$$, it follows from the second equation that $$a = 3y + 2$$ and from the first that $$b = \frac{3x - a}{2} = \frac{3x - 3y - 2}{2}$$.

The fixed length of the rod implies $$|AB|^2 = (a - b)^2 + \bigl(a + 2 - (-2)\bigr)^2 = (a - b)^2 + (a + 4)^2 = 64$$.

Substituting for $$a$$ and $$b$$ gives $$a - b = (3y + 2) - \frac{3x - 3y - 2}{2} = \frac{2(3y + 2) - 3x + 3y + 2}{2} = \frac{9y - 3x + 6}{2}$$ and $$a + 4 = 3y + 2 + 4 = 3y + 6$$.

Therefore $$(a - b)^2 + (a + 4)^2 = 64$$ becomes $$\left(\frac{9y - 3x + 6}{2}\right)^2 + (3y + 6)^2 = 64$$, which is equivalently written as $$\frac{(9y - 3x + 6)^2}{4} + (3y + 6)^2 = 64$$.

Noting that $$\frac{(9y - 3x + 6)^2}{4} = \frac{9(3y - x + 2)^2}{4}$$ and $$(3y + 6)^2 = 9(y + 2)^2$$, one obtains $$9(3y - x + 2)^2 + 36(y + 2)^2 = 256$$.

Expanding $$(3y - x + 2)^2 = 9y^2 + x^2 + 4 - 6xy + 12y - 4x$$ leads to the equation $$9(x^2 + 9y^2 + 4 - 6xy - 4x + 12y) + 36(y^2 + 4y + 4) = 256$$.

Collecting like terms gives $$9x^2 + 81y^2 + 36 - 54xy - 36x + 108y + 36y^2 + 144y + 144 = 256$$, or $$9x^2 + 117y^2 - 54xy - 36x + 252y + 180 = 256$$.

Rearranging, we have $$9(x^2 + 13y^2 - 6xy - 4x + 28y) + 180 = 256$$, so $$9(x^2 + 13y^2 - 6xy - 4x + 28y) = 76$$ and hence $$9(x^2 + 13y^2 - 6xy - 4x + 28y) - 76 = 0$$.

Comparing this with $$9(x^2 + \alpha y^2 + \beta xy + \gamma x + 28y) - 76 = 0$$ shows that $$\alpha = 13$$, $$\beta = -6$$ and $$\gamma = -4$$.

Finally, $$\alpha - \beta - \gamma = 13 - (-6) - (-4) = 13 + 6 + 4 = 23$$.

The correct answer is Option 3: 23.