Let the domains of the functions $$f(x) = \log_4 \log_3 \log_7 (8 - \log_2(x^2 + 4x + 5))$$ and $$g(x) = \sin^{-1}\left(\frac{7x+10}{x-2}\right)$$ be $$(\alpha, \beta)$$ and $$[\gamma, \delta]$$, respectively. Then $$\alpha^2 + \beta^2 + \gamma^2 + \delta^2$$ is equal to :

For every logarithm we must have a positive argument, and for $$\log_a(b)$$ to be positive ( $$a\gt1$$ ) the argument $$b$$ must exceed $$1$$.
We examine the functions layer by layer.

Case 1: Domain of $$f(x)=\log_{4}\!\Bigl[\;\log_{3}\!\bigl\{\log_{7}\!\bigl(8-\log_{2}(x^{2}+4x+5)\bigr)\bigr\}\Bigr]$$

Step 1 - innermost logarithm

$$x^{2}+4x+5=(x+2)^{2}+1\gt0$$ for all real $$x$$, so $$\log_{2}(x^{2}+4x+5)$$ exists for every real $$x$$.

Step 2 - argument of $$\log_{7}$$

Let $$y=8-\log_{2}(x^{2}+4x+5)$$. For $$\log_{7}(y)$$ to exist we need $$y\gt0$$, i.e.

$$8-\log_{2}(x^{2}+4x+5)\gt0 \;\Longrightarrow\; \log_{2}(x^{2}+4x+5)\lt8.$$

Step 3 - argument of $$\log_{3}$$ must be positive

Let $$z=\log_{7}(y)$$. For $$\log_{3}(z)$$ to be positive we require $$z\gt1$$ (because base $$3\gt1$$). Thus

$$\log_{7}\!\bigl(8-\log_{2}(x^{2}+4x+5)\bigr)\gt1 \;\Longrightarrow\; 8-\log_{2}(x^{2}+4x+5)\gt7.$$

Step 4 - simplifying the last inequality

$$8-\log_{2}(x^{2}+4x+5)\gt7 \;\Longrightarrow\; \log_{2}(x^{2}+4x+5)\lt1 \;\Longrightarrow\; x^{2}+4x+5\lt2.$$

Because $$x^{2}+4x+5=(x+2)^{2}+1$$, the above becomes

$$(x+2)^{2}+1\lt2 \;\Longrightarrow\; (x+2)^{2}\lt1 \;\Longrightarrow\; -1\lt x+2\lt1 \;\Longrightarrow\; -3\lt x\lt-1.$$

Hence the domain of $$f(x)$$ is $$(\alpha,\beta)=(-3,-1).$$ So $$\alpha=-3,\;\beta=-1.$$

Case 2: Domain of $$g(x)=\sin^{-1}\!\left(\dfrac{7x+10}{\,x-2\,}\right)$$

The argument of $$\sin^{-1}$$ must lie in $$[-1,1]$$ and the denominator cannot be zero.

Let $$t=\dfrac{7x+10}{x-2}$$. We need $$-1\le t\le1,\;x\ne2.$$

1) Inequality $$t\le1$$:

$$\dfrac{7x+10}{x-2}\le1 \;\Longrightarrow\; \dfrac{6(x+2)}{x-2}\le0.$$

This holds for $$-2\le x&lt2.$$

2) Inequality $$t\ge-1$$:

$$\dfrac{7x+10}{x-2}\ge-1 \;\Longrightarrow\; \dfrac{8(x+1)}{x-2}\ge0.$$

This is true for $$x\le-1$$ or $$x>2.$$

3) Intersection of (1) and (2):

$$[-2,2)\;\cap\;(-\infty,-1]\;=\;[-2,-1].$$

The endpoint $$x=-2$$ gives $$t=1$$ and $$x=-1$$ gives $$t=-1,$$ both allowed. Hence the domain of $$g(x)$$ is $$[\gamma,\delta]=[-2,-1]$$ with $$\gamma=-2,\;\delta=-1.$$

Case 3: Required sum

$$\alpha^2+\beta^2+\gamma^2+\delta^2 =(-3)^2+(-1)^2+(-2)^2+(-1)^2 =9+1+4+1 =15.$$

Therefore, $$\alpha^2+\beta^2+\gamma^2+\delta^2=15$$, which matches Option A.