vec$(T_{AC})=T_{AC}|(T_{\text{A}\text{A}\text{C}}^1,T_{\text{J}\text{A}\text{C}}(hat(j)),T_{2AC}(hat(k)))||$

$x_{AC}=x_C-x_{AC}=-0\mathrm{.}95-0=-0\mathrm{.}95m$

$y_{AC}=y_C-y_A=1\mathrm{.}3-0=1\mathrm{.}3m$

$z_{AC}=z_C-z_A=0-2\mathrm{.}6=-2\mathrm{.}6m$

$=>,$vec$(T_{AC})=|T_{AC}|[\frac{-0\mathrm{.}95(hat(i))+1\mathrm{.}3j-2\mathrm{.}6(hat(k))}{(\sqrt[\phantom{2}]{{0\mathrm{.}95}^2+{1\mathrm{.}3}^2+2\mathrm{.}62})}]=T_{AC}(-0\mathrm{.}31$hat$(i)+0\mathrm{.}42$hat$(j)-0\mathrm{.}85$hat$(k))$

$,3\mathrm{.}058$

${?}_{for{T}_{BD}}^{\text{S}\text{a}\text{m}\text{e}}$vec$(T_{BD})=|T_{BD}|(0\mathrm{.}707$hat$(j)-0\mathrm{.}707$hat$(k))$

vec$(T_{BE})=-T_{BE}$hat$(k)$

$=>$ NOw we Can have: $\{\begin{array}{c}{F}_x=0\\ {\displaystyle \underset{?}{\overset{?}{}}}{F}_y=0\\ {\displaystyle \underset{?}{\overset{?}{}}}{F}_z=0\end{array}$

${\displaystyle \underset{?}{\overset{?}{}}}{M}_{0x}=0,{\displaystyle \underset{?}{\overset{?}{}}}{M}_{0y}=0,{\displaystyle \underset{?}{\overset{?}{}}}{M}_{0z}=0$

${\displaystyle \underset{?}{\overset{?}{}}}{F}_x=0=>O_x-0\mathrm{.}31T_{AC}=0$

${\displaystyle \underset{?}{\overset{?}{}}}{r}_y=0=>O_y+0\mathrm{.}42T_{AC}+0\mathrm{.}70T_{T_{BD}}-5395\mathrm{.}5=0$

$S{F}_z=0=>O_z-0\mathrm{.}85T_{AC}-0\mathrm{.}707T_{BD}-T_{BE}=0$

$S_{BC}=0=>-0\mathrm{.}31T_{AC}2\mathrm{.}6-0\mathrm{.}707T_{B_D}2\mathrm{.}6+5395\mathrm{.}52\mathrm{.}6=0$

$ox$

distance $d{T}_{AC,x}$ from $x-ax.s,$