Solve the system of linear equations using the elimination method. \[ \begin{array}{l} \frac{1}{3} x-\frac{2}{3} y+z=-2 \\ \frac{1}{2} x-\frac{3}{4} y+z=-\frac{11}{4} \\ -2 x-y+z=4 \end{array} \]

Step 1 ：Multiply the first equation by 3, the second equation by 4, and the third equation by 1 to get rid of the fractions and make the coefficients of x the same in the first and second equations: \[3*(\frac{1}{3})x - 3*(\frac{2}{3})y + 3*z = 3*(-2)\] \[4*(\frac{1}{2})x - 4*(\frac{3}{4})y + 4*z = 4*(-\frac{11}{4})\] \[1*(-2)x - 1*y + 1*z = 1*4\]

Step 2 ：Simplify the equations to: \[x - 2y + 3z = -6\] \[2x - 3y + 4z = -11\] \[-2x - y + z = 4\]

Step 3 ：Subtract the first equation from the second to eliminate x: \[(2x - 3y + 4z) - (x - 2y + 3z) = -11 - (-6)\]

Step 4 ：Simplify the equation to: \[x - y + z = -5\]

Step 5 ：Subtract the first equation from the third to eliminate y: \[(-2x - y + z) - (x - y + z) = 4 - (-5)\]

Step 6 ：Simplify the equation to: \[-x = 9\] and solve for x to get \[x = -9\]

Step 7 ：Substitute x = -9 into the first equation: \[-9 - y + z = -5\] and simplify to \[-y + z = 4\] or \[y = z - 4\]

Step 8 ：Substitute x = -9 into the third equation: \[-2*(-9) - y + z = 4\] and simplify to \[18 - y + z = 4\] or \[y = z + 14\]

Step 9 ：Setting y = z - 4 and y = z + 14 equal to each other, we get: \[z - 4 = z + 14\] and solve for z to get \[z = -18\]

Step 10 ：Substitute z = -18 into y = z - 4: \[y = -18 - 4\] and solve for y to get \[y = -22\]

Step 11 ：So, the solution to the system of equations is \[\boxed{x = -9}\], \[\boxed{y = -22}\], \[\boxed{z = -18}\]