Step-by-step Solution:

1. Understand the Given Information We are given the following:

• $\overrightarrow a$ and $\overrightarrow b$ are unit vectors, so $|\overrightarrow a| = 1$ and $|\overrightarrow b| = 1$.
• $\overrightarrow a$ and $\overrightarrow b$ are orthogonal, so $\overrightarrow a \cdot \overrightarrow b = 0$.
• A vector $\overrightarrow c$ is defined as $\overrightarrow c = x\overrightarrow a + y\overrightarrow b + \overrightarrow a \times \overrightarrow b$.
• The magnitude of $\overrightarrow c$ is $| \overrightarrow c| = 2$.
• The vector $\overrightarrow c$ makes the same angle $\alpha$ with both $\overrightarrow a$ and $\overrightarrow b$.

2. Use the Angle Information to Relate x and y The cosine of the angle between two vectors $\overrightarrow u$ and $\overrightarrow v$ is given by $\cos \theta = \frac{\overrightarrow u \cdot \overrightarrow v}{|\overrightarrow u||\overrightarrow v|}$.

The angle between $\overrightarrow c$ and $\overrightarrow a$ is $\alpha$. So, $\cos \alpha = \frac{\overrightarrow c \cdot \overrightarrow a}{|\overrightarrow c||\overrightarrow a|}$ Let's compute the dot product $\overrightarrow c \cdot \overrightarrow a$: \overrightarrow c \cdot \overrightarrow a = (x\overrightarrow a + y\overrightarrow b + \overrightarrow a \times \overrightarrow b) \cdot \overrightarrow a $$$$ \overrightarrow c \cdot \overrightarrow a = x(\overrightarrow a \cdot \overrightarrow a) + y(\overrightarrow b \cdot \overrightarrow a) + (\overrightarrow a \times \overrightarrow b) \cdot \overrightarrow a Using the given properties:

• $\overrightarrow a \cdot \overrightarrow a = |\overrightarrow a|^2 = 1^2 = 1$
• $\overrightarrow b \cdot \overrightarrow a = 0$
• $(\overrightarrow a \times \overrightarrow b) \cdot \overrightarrow a = 0$ (since the scalar triple product with a repeated vector is zero). So, $\overrightarrow c \cdot \overrightarrow a = x(1) + y(0) + 0 = x$.

Substituting this back into the cosine formula: $\cos \alpha = \frac{x}{(2)(1)} = \frac{x}{2} \quad ...(1)$

Similarly, the angle between $\overrightarrow c$ and $\overrightarrow b$ is also $\alpha$. So, $\cos \alpha = \frac{\overrightarrow c \cdot \overrightarrow b}{|\overrightarrow c||\overrightarrow b|}$ Let's compute the dot product $\overrightarrow c \cdot \overrightarrow b$: \overrightarrow c \cdot \overrightarrow b = (x\overrightarrow a + y\overrightarrow b + \overrightarrow a \times \overrightarrow b) \cdot \overrightarrow b $$$$ \overrightarrow c \cdot \overrightarrow b = x(\overrightarrow a \cdot \overrightarrow b) + y(\overrightarrow b \cdot \overrightarrow b) + (\overrightarrow a \times \overrightarrow b) \cdot \overrightarrow b Using the given properties:

• $\overrightarrow a \cdot \overrightarrow b = 0$
• $\overrightarrow b \cdot \overrightarrow b = |\overrightarrow b|^2 = 1^2 = 1$
• $(\overrightarrow a \times \overrightarrow b) \cdot \overrightarrow b = 0$. So, $\overrightarrow c \cdot \overrightarrow b = x(0) + y(1) + 0 = y$.

Substituting this back into the cosine formula: $\cos \alpha = \frac{y}{(2)(1)} = \frac{y}{2} \quad ...(2)$

From equations (1) and (2), we have $\frac{x}{2} = \frac{y}{2}$, which implies $x = y$.

3. Use the Magnitude Information We are given $| \overrightarrow c| = 2$, which means $|\overrightarrow c|^2 = 4$. Let's compute $|\overrightarrow c|^2 = \overrightarrow c \cdot \overrightarrow c$: $|\overrightarrow c|^2 = (x\overrightarrow a + y\overrightarrow b + \overrightarrow a \times \overrightarrow b) \cdot (x\overrightarrow a + y\overrightarrow b + \overrightarrow a \times \overrightarrow b)$ The vectors $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow a \times \overrightarrow b$ are mutually orthogonal. This is because $\overrightarrow a \cdot \overrightarrow b = 0$, and by definition of the cross product, $\overrightarrow a \times \overrightarrow b$ is orthogonal to both $\overrightarrow a$ and $\overrightarrow b$. Therefore, the dot product of any two distinct vectors from this set is zero.

Expanding the dot product: |\overrightarrow c|^2 = x^2(\overrightarrow a \cdot \overrightarrow a) + y^2(\overrightarrow b \cdot \overrightarrow b) + (\overrightarrow a \times \overrightarrow b) \cdot (\overrightarrow a \times \overrightarrow b) + \text{terms with dot products of distinct vectors which are zero} $$$$ |\overrightarrow c|^2 = x^2|\overrightarrow a|^2 + y^2|\overrightarrow b|^2 + |\overrightarrow a \times \overrightarrow b|^2 We know $|\overrightarrow a|=1$ and $|\overrightarrow b|=1$. Also, $|\overrightarrow a \times \overrightarrow b| = |\overrightarrow a||\overrightarrow b|\sin\theta$, where $\theta$ is the angle between $\overrightarrow a$ and $\overrightarrow b$. Since $\overrightarrow a \cdot \overrightarrow b = 0$ and they are non-zero vectors, $\theta = 90^\circ$, so $\sin\theta = 1$. Thus, $|\overrightarrow a \times \overrightarrow b| = (1)(1)(1) = 1$.

Substituting these values: $|\overrightarrow c|^2 = x^2(1)^2 + y^2(1)^2 + (1)^2 = x^2 + y^2 + 1$ Given $|\overrightarrow c|^2 = 4$, we have: x^2 + y^2 + 1 = 4 $$$$ x^2 + y^2 = 3 \quad ...(3)

4. Solve for x and y We have a system of two equations:

1. $x = y$
2. $x^2 + y^2 = 3$

Substitute $y=x$ into the second equation: x^2 + x^2 = 3 $$$$ 2x^2 = 3 $$$$ x^2 = \frac{3}{2} Since $x=y$, we also have $y^2 = \frac{3}{2}$.

5. Calculate the Final Value We need to find the value of $8\cos^2\alpha$. From equation (1), we have $\cos \alpha = \frac{x}{2}$. Therefore, $\cos^2\alpha = \left(\frac{x}{2}\right)^2 = \frac{x^2}{4}$.

Substitute the value of $x^2 = \frac{3}{2}$: $\cos^2\alpha = \frac{3/2}{4} = \frac{3}{8}$

Finally, the value of $8\cos^2\alpha$ is: $8\cos^2\alpha = 8 \times \frac{3}{8} = 3$

The value of $8{\cos ^2}\alpha$ is 3.