Suppose $$28 - p$$, $$p$$, $$70 - \alpha$$, $$\alpha$$ are the coefficients of four consecutive terms in the expansion of $$(1 + x)^n$$. Then the value of $$2\alpha - 3p$$ equals

Four consecutive coefficients in $$(1+x)^n$$: $$\binom{n}{r-1} = 28-p$$, $$\binom{n}{r} = p$$, $$\binom{n}{r+1} = 70-\alpha$$, $$\binom{n}{r+2} = \alpha$$.

Using the property that consecutive binomial coefficients satisfy: $$\binom{n}{r-1} + \binom{n}{r} = \binom{n+1}{r}$$.

Sum of 1st and 2nd: $$(28-p) + p = 28 = \binom{n+1}{r}$$.

Sum of 3rd and 4th: $$(70-\alpha) + \alpha = 70 = \binom{n+1}{r+2}$$.

Sum of 2nd and 3rd: $$p + (70-\alpha) = \binom{n+1}{r+1}$$.

Now $$\binom{n+1}{r} = 28$$, $$\binom{n+1}{r+2} = 70$$.

Using $$\frac{\binom{n+1}{r+2}}{\binom{n+1}{r}} = \frac{(n+1-r)(n-r)}{(r+1)(r+2)} = \frac{70}{28} = \frac{5}{2}$$.

So $$2(n+1-r)(n-r) = 5(r+1)(r+2)$$.

Try $$n = 7, r = 1$$: $$2(7)(6) = 84$$, $$5(2)(3) = 30$$. No.

Try $$n = 7, r = 2$$: $$2(6)(5) = 60$$, $$5(3)(4) = 60$$. Yes!

Check: $$\binom{8}{2} = 28$$ ✓, $$\binom{8}{4} = 70$$ ✓.

$$\binom{8}{3} = 56 = p + (70-\alpha)$$, so $$p + 70 - \alpha = 56 \Rightarrow \alpha - p = 14$$.

Now $$\binom{7}{1} = 7 = 28 - p \Rightarrow p = 21$$. $$\binom{7}{2} = 21 = p$$ ✓.

$$\binom{7}{3} = 35 = 70 - \alpha \Rightarrow \alpha = 35$$. $$\binom{7}{4} = 35 = \alpha$$ ✓.

$$2\alpha - 3p = 70 - 63 = 7$$.

The answer is Option (1): $$\boxed{7}$$.