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Let $A = \begin{bmatrix}1&2&-3\\ 2&3&2\\ 3&-3&-4\end{bmatrix}$ and $B = \begin{bmatrix}-6&17&13\\ 14&5&-8\\ -15&9&-1\end{bmatrix}$. We need to find the product $AB$.
$AB = \begin{bmatrix}1&2&-3\\ 2&3&2\\ 3&-3&-4\end{bmatrix} \begin{bmatrix}-6&17&13\\ 14&5&-8\\ -15&9&-1\end{bmatrix} = \begin{bmatrix}(1)(-6)+(2)(14)+(-3)(-15)&(1)(17)+(2)(5)+(-3)(9)&(1)(13)+(2)(-8)+(-3)(-1)\\(2)(-6)+(3)(14)+(2)(-15)&(2)(17)+(3)(5)+(2)(9)&(2)(13)+(3)(-8)+(2)(-1)\\(3)(-6)+(-3)(14)+(-4)(-15)&(3)(17)+(-3)(5)+(-4)(9)&(3)(13)+(-3)(-8)+(-4)(-1)\end{bmatrix}$ $AB = \begin{bmatrix}-6+28+45&17+10-27&13-16+3\\-12+42-30&34+15+18&26-24-2\\-18-42+60&51-15-36&39+24+4\end{bmatrix} = \begin{bmatrix}67&0&0\\0&67&0\\0&0&67\end{bmatrix} = 67\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} = 67I$
The given system of linear equations is: $x+2y-3z=-4$ $2x+3y+2z=2$ $3x-3y-4z=11$ This can be written in matrix form as $AX = D$, where $A = \begin{bmatrix}1&2&-3\\ 2&3&2\\ 3&-3&-4\end{bmatrix}$, $X = \begin{bmatrix}x\\y\\z\end{bmatrix}$, and $D = \begin{bmatrix}-4\\2\\11\end{bmatrix}$.
Since $AB = 67I$, we have $A(\frac{1}{67}B) = I$. Therefore, $A^{-1} = \frac{1}{67}B = \frac{1}{67}\begin{bmatrix}-6&17&13\\ 14&5&-8\\ -15&9&-1\end{bmatrix}$.
We have $AX = D$, so $X = A^{-1}D$. $X = \frac{1}{67}\begin{bmatrix}-6&17&13\\ 14&5&-8\\ -15&9&-1\end{bmatrix} \begin{bmatrix}-4\\2\\11\end{bmatrix} = \frac{1}{67}\begin{bmatrix}(-6)(-4)+(17)(2)+(13)(11)\\(14)(-4)+(5)(2)+(-8)(11)\\(-15)(-4)+(9)(2)+(-1)(11)\end{bmatrix} = \frac{1}{67}\begin{bmatrix}24+34+143\\-56+10-88\\60+18-11\end{bmatrix} = \frac{1}{67}\begin{bmatrix}201\\-134\\67\end{bmatrix} = \begin{bmatrix}3\\-2\\1\end{bmatrix}$
Therefore, $x = 3$, $y = -2$, and $z = 1$.
Final Answer: x = 3, y = -2, z = 1