Solve the system below. Parameterize the solution set. \[ \begin{aligned}-x+2 y+5 z & =11 \\ -2 x+5 y+15 z & =25 \\ 3 x-3 y & =-24\end{aligned} \]

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}-x+2y+5z=11\\-2x+5y+15z=25\\3x-3y=-24\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=-11+2y+5z\\-2x+5y+15z=25\\3x-3y=-24\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left\{ \begin{array}{l}-2\left(-11+2y+5z\right)+5y+15z=25\\3\left(-11+2y+5z\right)-3y=-24\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}22+y+5z=25\\-33+3y+15z=-24\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}y=3-5z\\-33+3y+15z=-24\end{array}\right.\) - step5: Substitute the value of \(y:\) \(-33+3\left(3-5z\right)+15z=-24\) - step6: Simplify: \(-24=-24\) - step7: Substitute the value of \(y:\) \(x=-11+2\left(3-5z\right)+5z\) - step8: Simplify: \(x=-5-5z\) - step9: Calculate: \(\left(x,y,z\right) = \left(-5-5z,3-5z,z\right),z \in \mathbb{R}\) - step10: Alternative Form: \(\textrm{Infinitely many solutions}\) The system of equations provided has infinitely many solutions. To parameterize the solution set, we can express the variables in terms of a free variable. Let's denote \( z = t \) (where \( t \) is a parameter). We can then express \( x \) and \( y \) in terms of \( t \). 1. From the first equation: \[ -x + 2y + 5t = 11 \implies x = 2y + 5t - 11 \] 2. Substitute \( x \) into the second equation: \[ -2(2y + 5t - 11) + 5y + 15t = 25 \] Simplifying this: \[ -4y - 10t + 22 + 5y + 15t = 25 \] \[ y + 5t + 22 = 25 \implies y = 3 - 5t \] 3. Now substitute \( y \) back into the expression for \( x \): \[ x = 2(3 - 5t) + 5t - 11 \] \[ x = 6 - 10t + 5t - 11 = -5 - 5t \] Thus, the parameterized solution set can be expressed as: \[ \begin{aligned} x &= -5 - 5t, \\ y &= 3 - 5t, \\ z &= t, \end{aligned} \] where \( t \) is any real number.