a) Consider the linear system \( \left\{\begin{array}{c}x-y-3 z=3 \\ 2 x+z=0 \\ 2 y+7 z=c\end{array}\right. \) i) For what value of \( c \) does the system have a solution? ii) Solve the system at the value of \( c \) from part (i)?

To determine the value of \( c \) for which the system has a solution, we first express the second equation in terms of one variable. From the second equation \( 2x + z = 0 \), we can express \( z \) as \( z = -2x \). Next, substituting \( z = -2x \) into the first equation \( x - y - 3z = 3 \): \[ x - y - 3(-2x) = 3 \implies x - y + 6x = 3 \implies 7x - y = 3 \implies y = 7x - 3. \] Now, substituting \( z = -2x \) and \( y = 7x - 3 \) into the third equation \( 2y + 7z = c \): \[ 2(7x - 3) + 7(-2x) = c \implies 14x - 6 - 14x = c \implies -6 = c. \] Thus, for \( c = -6 \), the system will have a solution. Now, we will solve the system with \( c = -6 \). Substituting \( c = -6 \) back into the third equation, we don't need to change previous calculations. Using \( z = -2x \) and \( y = 7x - 3 \), we can pick any value for \( x \) to find the corresponding values of \( y \) and \( z \). Let’s choose \( x = 0 \): 1. From \( y = 7(0) - 3 = -3 \). 2. From \( z = -2(0) = 0 \). The solution to the system when \( c = -6 \) is: \[ \begin{align*} x & = 0, \\ y & = -3, \\ z & = 0. \end{align*} \] So, the final solution is \( (x, y, z) = (0, -3, 0) \).