We are given the system of equations:

$1. \ 2x - 6y - 3z = -26$ $2. \ -x - 12y + 3z = -11$ $3. \ 4x - 24y - 3z = -64$

To solve this system, let's proceed step by step.

Step 1: Simplify the equations, if possible

Equations 1 and 3 are already simplified. For Equation 2, we can multiply the whole equation by $-1$ to make the coefficient of $x$ positive:

$x + 12y - 3z = 11$

Step 2: Eliminate one variable

We will eliminate $z$ by adding or subtracting the equations.

First, let's add Equation 1 and Equation 3:

$(2x - 6y - 3z) + (4x - 24y - 3z) = -26 + (-64)$ $6x - 30y - 6z = -90$

Now divide the entire equation by 6: $x - 5y - z = -15 \quad \text{(Equation 4)}$

Step 3: Eliminate $z$ again with a different pair of equations

Now, let's add Equation 1 and the modified Equation 2 (after multiplying Equation 2 by 1):

$(2x - 6y - 3z) + (x + 12y - 3z) = -26 + 11$ $3x + 6y - 6z = -15$

Now divide this equation by 3: $x + 2y - 2z = -5 \quad \text{(Equation 5)}$

Step 4: Solve the new system

Now we have a simpler system of two equations in terms of $x$ and $y$ from Equation 4 and Equation 5:

1. $x - 5y - z = -15$
2. $x + 2y - 2z = -5$

We can now solve these two equations using substitution or elimination. Let's use substitution:

From Equation 1: $x - 5y = z - 15 \quad \text{or} \quad z = x - 5y + 15$

Substitue in...