Let `L` denots value of `cos^(2)(alpha - beta)` if `sin2alpha + sin2beta = (1)/(2), cos2alpha + cos2beta = (sqrt(3))/(2)` and `M` denotes value of `(1)/(log_(xy)xyz) + (1)/(log_(yz)xyz) + (1)/(log_(zx)xyz)` then `16L^(2) + M^(2)` is

Explanation

To solve the problem step by step, we will first find the value of \( L \) and then the value of \( M \), and finally compute \( 16L^2 + M^2 \).

### Step 1: Finding \( L = \cos^2(\alpha - \beta) \)

We are given:
1. \( \sin 2\alpha + \sin 2\beta = \frac{1}{2} \)
2. \( \cos 2\alpha + \cos 2\beta = \frac{\sqrt{3}}{2} \)

We can use the identity:
\[ \sin^2 x + \cos^2 x = 1 \]
Squaring both equations and adding them, we have:
\[ (\sin 2\alpha + \sin 2\beta)^2 + (\cos 2\alpha + \cos 2\beta)^2 = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \]

Calculating the right-hand side:
\[ \left(\frac{1}{2}\right)^2 = \frac{1}{4}, \quad \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]
Thus,
\[ \frac{1}{4} + \frac{3}{4} = 1 \]

Now, expanding the left-hand side:
\[ (\sin 2\alpha)^2 + 2\sin 2\alpha \sin 2\beta + (\sin 2\beta)^2 + (\cos 2\alpha)^2 + 2\cos 2\alpha \cos 2\beta + (\cos 2\beta)^2 \]
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ 1 + 1 + 2(\sin 2\alpha \sin 2\beta + \cos 2\alpha \cos 2\beta) = 1 \]
This simplifies to:
\[ 2 + 2 \cos(2\alpha - 2\beta) = 1 \]
Thus:
\[ 2 \cos(2\alpha - 2\beta) = -1 \implies \cos(2\alpha - 2\beta) = -\frac{1}{2} \]
This implies:
\[ 2\alpha - 2\beta = \frac{2\pi}{3} + 2k\pi \quad \text{or} \quad 2\alpha - 2\beta = \frac{4\pi}{3} + 2k\pi \]

Now, we can find \( \cos^2(\alpha - \beta) \):
Using the double angle formula:
\[ \cos(2\theta) = 2\cos^2(\theta) - 1 \]
Setting \( \theta = \alpha - \beta \):
\[ \cos(2(\alpha - \beta)) = -\frac{1}{2} \implies 2\cos^2(\alpha - \beta) - 1 = -\frac{1}{2} \]
Solving for \( \cos^2(\alpha - \beta) \):
\[ 2\cos^2(\alpha - \beta) = \frac{1}{2} \implies \cos^2(\alpha - \beta) = \frac{1}{4} \]
Thus, \( L = \frac{1}{4} \).

### Step 2: Finding \( M \)

We need to evaluate:
\[ M = \frac{1}{\log_{xy}(xyz)} + \frac{1}{\log_{yz}(xyz)} + \frac{1}{\log_{zx}(xyz)} \]

Using the change of base formula:
\[ \frac{1}{\log_a b} = \log_b a \]
We can rewrite \( M \):
\[ M = \log_{xyz}(xy) + \log_{xyz}(yz) + \log_{xyz}(zx) \]
Using the property of logarithms:
\[ M = \log_{xyz}(xy \cdot yz \cdot zx) = \log_{xyz}(x^2y^2z^2) = 2 \log_{xyz}(xyz) = 2 \]

### Step 3: Calculating \( 16L^2 + M^2 \)

Now substituting the values of \( L \) and \( M \):
\[ 16L^2 + M^2 = 16 \left(\frac{1}{4}\right)^2 + 2^2 \]
Calculating \( L^2 \):
\[ L^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \]
Thus:
\[ 16L^2 = 16 \cdot \frac{1}{16} = 1 \]
And:
\[ M^2 = 2^2 = 4 \]
Finally:
\[ 16L^2 + M^2 = 1 + 4 = 5 \]

### Final Answer
The value of \( 16L^2 + M^2 \) is \( \boxed{5} \).