<prompt> Katika pembetatu ABC, E iko kwenye overlineAC kiasi kwamba AE:EC 2:1, na F iko kwenye overlineAB kiasi kwamba AF:FB 1:4. Acha P iwe makutano ya overlineBE na overlineCF.asyukubwa wa kitengo sentimita 0.8;jozi A, B, C, D, E, F, P;A 1,4;B 0,0;C 6,0;E interpA,C,23;F interpA,B,15;P kiendeleziB,E,C,F;kuchoraA--B--C--mzunguko;kuchoraB--E;kuchoraC--F;leboA, A, N;leboB, B, SW;leboC, C, SE;leboE, E, NE;leboF, F, W;leboP, P, S;asyKishaarrow overrightarrowP x overrightarrowA y overrightarrowB z overrightarrowC,ambapo x, y, na z ni vitu visivyobadilika hivi kwamba x y z 1. Ingiza x,y,z iliyoagizwa mara tatu.  <response>: Kutoka kwa habari iliyotolewa,overrightarrowE frac13 overrightarrowA frac23 overrightarrowCnaoverrightarrowF frac45 overrightarrowA frac15 overrightarrowB.Kutenga overrightarrowA katika kila mlinganyo, tunapataoverrightarrowA 3 overrightarrowE - 2 overrightarrowC frac5 overrightarrowF - overrightarrowB4.Kisha 12 overrightarrowE - 8 arrowoverrightarrowC 5 overrightarrowF - overrightarrowB, hivyo 12 overrightarrowE overrightarrowB 5 overrightarrowF 8 overrightarrowC , aufrac1213 overrightarrowE frac113 overrightarrowB frac513 overrightarrowF frac8101 13 mshale uliopitilizaC.Kwa vile viambatanishi vya pande zote mbili za mlingano vinaongeza hadi 1, vekta iliyo upande wa kushoto iko kwenye mstari BE, na vekta iliyo upande wa kulia iko kwenye mstari CF. Kwa hivyo, vekta hii ya kawaida ni overrightarrowP. KishaanzaalignoverrightarrowP frac1213 overrightarrowE frac113 overrightarrowB frac1213 left frac13 overrightarrowA frac23 overrightarrowC kulia frac13 13 mshale wa kuliaB frac413 overrightarrowA frac113 overrightarrowB frac813 overrightarrowC.endalignHivyo, x,y,z left frac413, frac113, frac813 kulia. 