To solve this system using Cramer's Rule, let’s first write down the equations clearly:

1. $x + 3y + z = -1$
2. $x + 2y + 3z = 4$
3. $x + 8y + 5z = -7$

Cramer's Rule uses determinants to solve systems of linear equations of the form:

$Ax = B$

where $A$ is the matrix of coefficients, $x$ is the vector of variables, and $B$ is the constants vector.

Step 1: Matrix Setup

Let’s define the coefficient matrix $A$ and the constant vector $B$:

$A = \begin{bmatrix} 1 & 3 & 1 \\ 1 & 2 & 3 \\ 1 & 8 & 5 \end{bmatrix}, \quad B = \begin{bmatrix} -1 \\ 4 \\ -7 \end{bmatrix}$

Step 2: Determinant of $A$

To find $x$, $y$, and $z$ using Cramer's Rule, we first calculate $\det(A)$.

$\det(A) = \begin{vmatrix} 1 & 3 & 1 \\ 1 & 2 & 3 \\ 1 & 8 & 5 \end{vmatrix}$

Expanding along the first row:

$\det(A) = 1 \cdot \begin{vmatrix} 2 & 3 \\ 8 & 5 \end{vmatrix} - 3 \cdot \begin{vmatrix} 1 & 3 \\ 1 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 \\ 1 & 8 \end{vmatrix}$

Calculating each minor:

1. $\begin{vmatrix} 2 & 3 \\ 8 & 5 \end{vmatrix} = (2 \cdot 5) - (3 \cdot 8) = 10 - 24 = -14$
2. $\begin{vmatrix} 1 & 3 \\ 1 & 5 \end{vmatrix} = (1 \cdot 5) - (3 \cdot 1) = 5 - 3 = 2$
3. $\begin{vmatrix} 1 & 2 \\ 1 & 8 \end{vmatrix} = (1 \cdot 8) - (2 \cdot 1) = 8 - 2 = 6$

So,

$\det(A) = 1 \cdot (-14) - 3 \cdot 2 + 1 \cdot 6 = -14 - 6 + 6 = -14$

Step 3: Determinants for $x$, $y$, and $z$

Now, we replace each column of $A$ with $B$ to form matrices $A_x$, $A_y$, and $A_z$ and compute their determinants.

$\det(A_x)$:

$A_x = \begin{bmatrix} -1 & 3 & 1 \\ 4 & 2 & 3 \\ -7 & 8 & 5 \end{bmatrix}$

Expanding along the first row:

$\det(A_x) = -1 \cdot \begin{vmatrix} 2 & 3 \\ 8 & 5 \end{vmatrix} - 3 \cdot \begin{vmatrix} 4 & 3 \\ -7 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} 4 & 2 \\ -7 & 8 \end{vmatrix}$

Using the minors computed:

1. $\begin{vmatrix} 2 & 3 \\ 8 & 5 \end{vmatrix} = -14$
2. $\begin{vmatrix} 4 & 3 \\ -7 & 5 \end{vmatrix} = (4 \cdot 5) - (3 \cdot -7) = 20 + 21 = 41$
3. $\begin{vmatrix} 4 & 2 \\ -7 & 8 \end{vmatrix} = (4 \cdot 8) - (2 \cdot -7) = 32 + 14 = 46$

Thus,

$\det(A_x) = (-1)(-14) - 3(41) + 1(46) = 14 - 123 + 46 = -63$

$\det(A_y)$:

$A_y = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 4 & 3 \\ 1 & -7 & 5 \end{bmatrix}$

Expanding and solving similarly, we find $\det(A_y)$.

$\det(A_z)$:

[ A_z = \begin{bmatrix}