Given equations when written in the form of A*F=B, we get

$\begin{bmatrix} 1 &3 &4 \\-1 &2 &-4 \\2 &-3 &2 \end{bmatrix} \begin{bmatrix} x\\y \\z \end{bmatrix} = \begin{bmatrix} 2\\2 \\2\end{bmatrix} $

 Generating an Auxillary matrix, we get,

$\begin{bmatrix} \begin{matrix} l_{11} &u_{12} &u_{13} \\l_{21} &l_{22} & u_{23} \\ l_{31} & l_{32} & l_{33} \end{matrix}& \Bigg | \begin{matrix} v_1\\v_2 \\v_3 \end{matrix}\end{bmatrix}$

Now, let's find step by step the value of all above unknown factors using Crout's 

Step1: Finding  $L_{11}, \ l_{12}$ and $l_{31}$

$l_{11} = a_{11}; \ l_{21} = a_{21}; \ l_{31} = a_{31}$

So, $l_{11} =1; \ l_{21}= -1; \ l_{31} = 2$

Hence,

$\begin{bmatrix} \begin{matrix} 1 &u_{12} &u_{13} \\-1 &l_{22} & u_{23} \\ 2 & l_{32} & l_{33} \end{matrix}& \Bigg| \begin{matrix} v_1\\v_2 \\v_3 \end{matrix}\end{bmatrix}$

Step 2: Finding $u_{12}, u_{13},$ and $v_1$

$u_{12}= \dfrac {a_{12}}{l_{11}} = 3$

$u_{13}= \dfrac {a_{12}}{l_{11}} = 4$

$v_1= \dfrac {b_1}{l_{11}} = 2$

Hence the matrix now looks

$\begin{bmatrix} \begin{matrix} 1 &3 &4 \\-1 &l_{22} & u_{23} \\ 2 & l_{32} & l_{33} \end{matrix}& \Bigg| \begin{matrix}2\\v_2 \\v_3 \end{matrix}\end{bmatrix}$

Step 3: Finding $l_{22}$ and $l_{32}$

$l_{22} = a_{22} - l_{21} * u_{12} = 2 -1 (3) = 5$

$l_{32} =a_{32} - l_{31} * u_{12} = -3-(2) (3) = -9$

So, the matrix is,

$\begin{bmatrix} \begin{matrix} 1 &3 &4 \\-1 &5 & u_{23} \\ 2 & -9 & l_{33} \end{matrix}& \Bigg| \begin{matrix}2\\v_2 \\v_3 \end{matrix}\end{bmatrix}$

Step 4: Finding $u_{23}$ and $v_2$

$u_{23} = \dfrac {a_{23}-l_{21}*u_{13}}{l_{22}} = \dfrac {-4-(1)(4)}{2} = 0$

$v_2= \dfrac {b_2 - l_{21}*v_1}{l_{22}}= \dfrac {2-(-1)(2)}{2} =2$

Hence, the matrix is,

$\begin{bmatrix} \begin{matrix} 1 &3 &4 \\-1 &5 & 0\\ 2 & -9 & l_{33} \end{matrix}& \Bigg| \begin{matrix}2\\2\\v_3 \end{matrix}\end{bmatrix}$

Step 5: Finding $l_{33} $ and $v_3$

$l_{33} = a_{33} - (l_{31}*u_{13}+ l_{32}*u_{23}) = 2 - \big [2*4+(-9)(0) \big] = -6$

$ v_{3} = \dfrac {b_3 - (l_{31}* v_1 + l_{32}*v_2)}{l_{33}} = \dfrac {2\big [2*2 = (-9) (2) \big]}{-6} = \dfrac {2-(-14)}{6} = \dfrac {8}{3}$

So, the final auxiliary matrix is,

$ \begin{bmatrix} \begin{matrix} 1 &3 &4 \\-1 &5 & 0 \\ 2 & -9 & -6 \end{matrix}& \Bigg| \begin{matrix}2\\2 \\ \frac {8}{3} \end{matrix}\end{bmatrix}$

Considering the unitary matrix,

U*F=V

$\begin{bmatrix} 1 &3 &4 \\0 &1 &0 \\0 &0 &1 \end{bmatrix} \begin{bmatrix} x\\y \\z\end{bmatrix} = \begin{bmatrix} 2\\2\\ \frac {8}{3}\end{bmatrix}$

On solving,

$x+3y = 4z = 2$

$y=2$ and

$z=\dfrac {8}{3}$

Substituting y and z in the first equation, we get,

$x=-\dfrac {44}{3}$

Hence, the values are $x= - \dfrac {44}{3}; \ y=2 $ and $z=\dfrac {8}{3}$