If the coefficient of x in the expansion of $$(ax^{2}+bx+c)(1-2x)^{26}$$. is - 56 and the coefficients of $$x^{2}\text{ and }x^{3}$$ are both zero, then a + b + c is equal to:

Coefficient of $$x$$ in $$(ax^2+bx+c)(1-2x)^{26} = -56$$, coefficients of $$x^2$$ and $$x^3$$ are both zero.

$$(1-2x)^{26} = \sum_{r=0}^{26}\binom{26}{r}(-2x)^r$$

Coeff of $$x^0$$ in $$(1-2x)^{26}$$: 1

Coeff of $$x^1$$: $$-52$$

Coeff of $$x^2$$: $$\binom{26}{2}(4) = 325 \times 4 = 1300$$

Coeff of $$x^3$$: $$\binom{26}{3}(-8) = 2600 \times (-8) = -20800$$

Coefficient of x in product: $$b \times 1 + c \times (-52) = b - 52c = -56$$ ... (1)

Coefficient of x² in product: $$a \times 1 + b(-52) + c(1300) = a - 52b + 1300c = 0$$ ... (2)

Coefficient of x³ in product: $$a(-52) + b(1300) + c(-20800) = -52a + 1300b - 20800c = 0$$ ... (3)

From (3): $$-52a + 1300b - 20800c = 0 \Rightarrow a - 25b + 400c = 0$$ ... (3')

From (2): $$a = 52b - 1300c$$. Sub in (3'): $$52b - 1300c - 25b + 400c = 0 \Rightarrow 27b = 900c \Rightarrow b = \frac{100c}{3}$$.

From (1): $$\frac{100c}{3} - 52c = -56 \Rightarrow \frac{100c - 156c}{3} = -56 \Rightarrow \frac{-56c}{3} = -56 \Rightarrow c = 3$$.

$$b = 100, a = 52(100) - 1300(3) = 5200 - 3900 = 1300$$.

$$a + b + c = 1300 + 100 + 3 = 1403$$.

The answer is Option 2: 1403.