Domain, Codomain and Range

To define the domain, let's analyse the argument and the base of the logarithmic function $$f(x)$$:

Base: $$10x^{2}-17x+7>0$$

Which gives $$x>\dfrac{17+\sqrt{9}}{20}$$ and $$x<\dfrac{17-\sqrt{9}}{20}$$  or  $$x>1$$ and $$x<0.7$$

Also, the base $$10x^{2}-17x+7\neq 1$$ gives,

$$10x^2-17x+6\neq 0$$  or  $$x\neq \dfrac{17\pm \sqrt{49}}{20}$$

Thus, $$x\neq 1.2$$ and $$x\neq 0.5$$

Of these, $$1.2$$ only falls in the given range, and will have to be excluded.

Similarly,

Argument: $$18x^{2}-11x+1>0$$

Which gives $$x>\dfrac{11+\sqrt{49}}{36}$$ and $$x<\dfrac{11-\sqrt{49}}{36}$$  or  $$x>0.5$$ and $$x<0.\overline{1}$$, where $$0.\overline{1} = \dfrac{1}{9}$$

Thus, the combined domain is;

$$(-\infty, 0.\overline{1}) \cup (0.5, 0.7) \cup (1, \infty) - {1.2}$$

Hence, $$a=\dfrac{1}{9}$$, $$b=0.5$$, $$c=0.7$$, $$d=1$$, and $$e=1.2$$

Which gives 
$$90(a + b + c + d + e) = 10 + 45 + 63+ 90 + 108 = \boxed{316}$$

We need to find the domain of $$f(x) = \log_x(1 - \log_4(x^2 - 9x + 18))$$.

For $$\log_x(\cdot)$$ to be defined, we need $$x \gt 0$$, $$x \neq 1$$.

For $$\log_4(x^2 - 9x + 18)$$ to be defined, we need $$x^2 - 9x + 18 \gt 0$$, i.e., $$(x-3)(x-6) \gt 0$$, so $$x \lt 3$$ or $$x \gt 6$$.

The argument of the outer logarithm must be positive: $$1 - \log_4(x^2 - 9x + 18) \gt 0$$, which gives $$\log_4(x^2 - 9x + 18) \lt 1$$, so $$x^2 - 9x + 18 \lt 4$$, i.e., $$x^2 - 9x + 14 \lt 0$$, giving $$(x-2)(x-7) \lt 0$$, so $$2 \lt x \lt 7$$.

Combining all conditions: $$x \gt 0$$, $$x \neq 1$$, ($$ x \lt 3$$ or $$x \gt 6$$), and $$2 \lt x \lt 7$$:

From $$x \lt 3$$ and $$2 \lt x \lt 7$$: we get $$2 \lt x \lt 3$$. Since $$x \neq 1$$ is automatically satisfied, this gives $$(2, 3)$$.

From $$x \gt 6$$ and $$2 \lt x \lt 7$$: we get $$6 \lt x \lt 7$$, giving $$(6, 7)$$.

So $$(\alpha, \beta) = (2, 3)$$ and $$(\gamma, \delta) = (6, 7)$$.

Therefore, $$\alpha + \beta + \gamma + \delta = 2 + 3 + 6 + 7 = 18$$.

Hence, the correct answer is Option A.

Given the domain of the function $$f(x) = \cos^{-1}\left(\frac{4x+5}{3x-7}\right)$$ is $$[\alpha, \beta]$$. The domain of the inverse cosine function is $$[-1, 1]$$, we get:

$$-1 \le \frac{4x+5}{3x-7} \le 1$$

From the inequality $$\frac{4x+5}{3x-7} \le 1$$:

$$\frac{4x+5}{3x-7} - 1 \le 0 \implies \frac{4x+5 - (3x-7)}{3x-7} \le 0 \implies \frac{x+12}{3x-7} \le 0$$

The critical points are $$x = -12$$ and $$x = \frac{7}{3}$$. This gives:

$$x \in \left[-12, \frac{7}{3}\right)$$

From the inequality $$\frac{4x+5}{3x-7} \ge -1$$:

$$\frac{4x+5}{3x-7} + 1 \ge 0 \implies \frac{4x+5 + (3x-7)}{3x-7} \ge 0 \implies \frac{7x-2}{3x-7} \ge 0$$

The critical points are $$x = \frac{2}{7}$$ and $$x = \frac{7}{3}$$. This gives:

$$x \in \left(-\infty, \frac{2}{7}\right] \cup \left(\frac{7}{3}, \infty\right)$$

Taking the intersection of both intervals gives the domain of $$f(x)$$:

$$\text{Domain of } f(x) = \left[-12, \frac{2}{7}\right]$$

Comparing this with $$[\alpha, \beta]$$, we get:

$$\alpha = -12, \quad \beta = \frac{2}{7}$$

$$7(\alpha + \beta) = 7\left(-12 + \frac{2}{7}\right) = -84 + 2 = -82$$

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Given the domain of the function $$g(x) = \log_2(2 - 6\log_{27}(2x+5))$$ is $$(\gamma, \delta)$$. For the log functions to be defined, their arguments must be strictly positive.

From the inner log argument:

$$2x + 5 > 0 \implies x > -\frac{5}{2}$$

From the outer log argument:

$$2 - 6\log_{27}(2x+5) > 0 \implies 6\log_{27}(2x+5) < 2 \implies \log_{27}(2x+5) < \frac{1}{3}$$

Converting to exponential form:

$$2x + 5 < (27)^{\frac{1}{3}} \implies 2x + 5 < 3 \implies 2x < -2 \implies x < -1$$

Taking the intersection of both conditions gives the domain of $$g(x)$$:

$$\text{Domain of } g(x) = \left(-\frac{5}{2}, -1\right)$$

Comparing this with $$(\gamma, \delta)$$, we get:

$$\gamma = -\frac{5}{2}, \quad \delta = -1$$

$$4(\gamma + \delta) = 4\left(-\frac{5}{2} - 1\right) = 4\left(-\frac{7}{2}\right) = -14$$

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Now, substituting these values into the required expression:

$$\left|7(\alpha + \beta) + 4(\gamma + \delta)\right| = |-82 - 14| = |-96| = 96$$

Therefore, the final value is equal to 96.