Solve the system by any method. \[ \begin{array}{r} x+4 y-z=3 \\ 3 x+9 y+z+3 w=6 \\ 2 x+5 y+2 z+3 w=4 \end{array} \] Select the correct answer, and if necessary, fill in the answer boxes to complete your choice. A. The system has exactly one solution ( \( \square \) , \( \square \) \( \square \) , \( \square \) B. This system has infinitely many solutions of the form \( \square \) , ,, , w), where \( w \) is any real number. (Type expressions using \( w \) as the variable.) C. The system has no solution.

To solve this system of equations, we can use the method of elimination or substitution. Let's simplify the equations provided. 1. The first equation is: \[ x + 4y - z = 3 \quad (1) \] 2. The second equation simplifies (after rearranging) to: \[ 3x + 9y + z + 3w = 6 \quad (2) \] 3. The third equation can also be rearranged to: \[ 2x + 5y + 2z + 3w = 4 \quad (3) \] Next, let's express \( z \) from Equation (1): \[ z = x + 4y - 3 \] Now that we have \( z \), we can substitute it into Equations (2) and (3) to eliminate \( z \). Substituting into Equation (2): \[ 3x + 9y + (x + 4y - 3) + 3w = 6 \] This simplifies to: \[ 4x + 13y + 3w - 3 = 6 \quad \implies 4x + 13y + 3w = 9 \quad (4) \] Substituting into Equation (3): \[ 2x + 5y + 2(x + 4y - 3) + 3w = 4 \] This simplifies to: \[ 2x + 5y + 2x + 8y - 6 + 3w = 4 \quad \implies 4x + 13y + 3w - 6 = 4 \quad \implies 4x + 13y + 3w = 10 \quad (5) \] Now, we have a simplified system of two equations: \[ 4x + 13y + 3w = 9 \quad (4) \] \[ 4x + 13y + 3w = 10 \quad (5) \] Observe that Equations (4) and (5) are contradictory (the left sides are equal, but the right sides are different). This indicates that there is no solution to this system. Thus, the conclusion is: C. The system has no solution.