Solve the system of linear equations. \[ \left\{\begin{array}{rr}x+2 y-7 z= & -4 \\ 2 x+3 y+z= & 5 \\ 3 x+7 y-36 z= & -25\end{array}\right. \]

To solve the given system of linear equations, we can use the method of substitution or elimination. Here's how to approach it step by step: We have the system: 1. \( x + 2y - 7z = -4 \) (Equation 1) 2. \( 2x + 3y + z = 5 \) (Equation 2) 3. \( 3x + 7y - 36z = -25 \) (Equation 3) Let's solve this system. First, let's express \( x \) from Equation 1: \[ x = -4 - 2y + 7z \quad (4) \] Now substitute Equation (4) into Equations 2 and 3: Substituting into Equation 2: \[ 2(-4 - 2y + 7z) + 3y + z = 5 \] \[ -8 - 4y + 14z + 3y + z = 5 \] \[ -8 - y + 15z = 5 \] \[ -y + 15z = 13 \quad (5) \] Now substitute Equation (4) into Equation 3: \[ 3(-4 - 2y + 7z) + 7y - 36z = -25 \] \[ -12 - 6y + 21z + 7y - 36z = -25 \] \[ -12 + y - 15z = -25 \] \[ y - 15z = -13 \quad (6) \] Now we have the new system of equations (5) and (6): 1. \(-y + 15z = 13\) (Equation 5) 2. \(y - 15z = -13\) (Equation 6) Adding Equations (5) and (6): \[ (-y + 15z) + (y - 15z) = 13 - 13 \] \[ 0 = 0 \] This shows that the two equations are dependent, confirming they represent the same line. Therefore, we can solve for \( y \) in terms of \( z \). From Equation (5): \[ y = 15z + 13 \quad (7) \] Now substituting Equation (7) back into Equation (4): \[ x = -4 - 2(15z + 13) + 7z \] \[ x = -4 - 30z - 26 + 7z \] \[ x = -30 - 23z \] Now we can express the solution as parametric equations in terms of \( z \): \[ x = -30 - 23z, \quad y = 15z + 13, \quad z = z \] Thus, the general solution is: \[ \boxed{(x, y, z) = (-30 - 23t, 15t + 13, t) \text{ for any } t \in \mathbb{R}} \]