Given linear programming problem is

7x + 7y - 7z = 2

-x + 11y + 7z = 1

11x + 5y + 7z = 0

It's matrix form is AX = B.

Where, A = \(\begin{bmatrix}7&7&-7\\-1&11&7\\11&5&7\end{bmatrix},\) x = \(\begin{bmatrix}x\\y\\z\end{bmatrix},\) B = \(\begin{bmatrix}2\\1\\0\end{bmatrix}\)

D = |A| = \(\begin{vmatrix}7&7&-7\\-1&11&7\\11&5&7\end{vmatrix}\) = \(7\begin{vmatrix}11&7\\5&7\end{vmatrix}-7\begin{vmatrix}-1&7\\11&7\end{vmatrix}\)\(-7\begin{vmatrix}-1&11\\11&5\end{vmatrix}\)

= 7(77 - 35) - 7(-7 - 77) - 7(-5 - 121)

 = 7 x 42 - 7x - 84 - 7x - 126

 = 294 + 588 + 882

 = 1764

D_x = \(\begin{vmatrix}2&7&-7\\1&11&7\\0&5&7\end{vmatrix}\)  = \(2\begin{vmatrix}11&7\\5&7\end{vmatrix}-7\begin{vmatrix}1&7\\0&7\end{vmatrix}\)\(-7\begin{vmatrix}1&11\\0&5\end{vmatrix}\)

 = 2(77 - 35) - 7(7 - 0) - 7(5 - 0)

 = 2 x 42 - 49 - 35

 = 84 - 84 = 0

D_y = \(\begin{vmatrix}7&2&-7\\-1&1&7\\11&0&7\end{vmatrix}\) =  \(7\begin{vmatrix}1&7\\0&7\end{vmatrix}-2\begin{vmatrix}-1&7\\11&7\end{vmatrix}\)\(-7\begin{vmatrix}-1&1\\11&0\end{vmatrix}\)

 = 7( 7 - 0) - 2(-7 - 77) - 7(0 - 11)

 = 49 - 2x - 84 + 77

 = 49 + 168 + 77

 = 294

D_z = \(\begin{vmatrix}7&7&2\\-1&11&1\\11&5&0\end{vmatrix}\) =  =  \(7\begin{vmatrix}11&1\\5&0\end{vmatrix}-7\begin{vmatrix}-1&1\\11&0\end{vmatrix}\)\(+7\begin{vmatrix}-1&11\\11&5\end{vmatrix}\)

 = 7(0 - 5) - 7(0 - 11) + 2(-5 + 121)

 = -35 + 77 - 252

 = -287 + 77

 = -210

x = \(\frac{D_x}{D}=\frac{0}{1764} = 0,\) 

y = \(\frac{D_y}{D}=\frac{294}{1784} =\frac16\) 

z = \(\frac{D_z}{D}=\frac{-210}{1764} =\frac{-10}{84}=\frac{-5}{42}\) 

Hence, solution of given linear programming is 

x = 0, y = 1/6 and z = -5/42