Let the altitudes be $\overline{AD},$ $\overline{BE},$ and $\overline{CF}.$

[asy]
unitsize(0.6 cm);

pair A, B, C, D, E, F, H;

B = (0,0);
C = (8,0);
A = intersectionpoint(arc(B,5,0,180),arc(C,7,0,180));
H = orthocenter(A,B,C);
D = (A + reflect(B,C)*(A))/2;
E = (B + reflect(C,A)*(B))/2;
F = (C + reflect(A,B)*(C))/2;

draw(A--B--C--cycle);
draw(A--D,dashed);
draw(B--E,dashed);
draw(C--F,dashed);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NE);
label("$F$", F, NW);
label("$H$", H, SE, UnFill);

dot(H);
[/asy]

By the Law of Cosines,
\begin{align*}
\cos A &= \frac{5^2 + 7^2 - 8^2}{2 \cdot 5 \cdot 7} = \frac{1}{7}, \\
\cos B &= \frac{5^2 + 8^2 - 7^2}{2 \cdot 5 \cdot 8} = \frac{1}{2}, \\
\cos C &= \frac{7^2 + 8^2 - 5^2}{2 \cdot 7 \cdot 8} = \frac{11}{14}.
\end{align*}Then $BD = AB \cos B = \frac{5}{2}$ and $CD = AC \cos C = \frac{11}{2},$ so
\[\overrightarrow{D} = \frac{11}{16} \overrightarrow{B} + \frac{5}{16} \overrightarrow{C}.\]Also, $AE = AB \cos A = \frac{5}{7}$ and $CE = BC \cos C = \frac{44}{7},$ so
\[\overrightarrow{E} = \frac{44}{49} \overrightarrow{A} + \frac{5}{49} \overrightarrow{C}.\]Isolating $\overrightarrow{C}$ in these equations, we obtain
\[\overrightarrow{C} = \frac{16 \overrightarrow{D} - 11 \overrightarrow{B}}{5} = \frac{49 \overrightarrow{E} - 44 \overrightarrow{A}}{5}.\]Then $16 \overrightarrow{D} - 11 \overrightarrow{B} = 49 \overrightarrow{E} - 44 \overrightarrow{A},$ so $16 \overrightarrow{D} + 44 \overrightarrow{A} = 49 \overrightarrow{E} + 11 \overrightarrow{B},$ or
\[\frac{16}{60} \overrightarrow{D} + \frac{44}{60} \overrightarrow{A} = \frac{49}{60} \overrightarrow{E} + \frac{11}{60} \overrightarrow{B}.\]Since the coefficients on both sides of the equation add up to 1, the vector on the left side lies on line $AD,$ and the vector on the right side lies on line $BE.$  Therefore, this common vector is $\overrightarrow{H}.$  Then
\begin{align*}
\overrightarrow{H} &= \frac{49}{60} \overrightarrow{E} + \frac{11}{60} \overrightarrow{B} \\
&= \frac{49}{60} \left( \frac{44}{49} \overrightarrow{A} + \frac{5}{49} \overrightarrow{C} \right) + \frac{11}{60} \overrightarrow{B} \\
&= \frac{11}{15} \overrightarrow{A} + \frac{11}{60} \overrightarrow{B} + \frac{1}{12} \overrightarrow{C}.
\end{align*}Thus, $(x,y,z) = \left( \frac{11}{15}, \frac{11}{60}, \frac{1}{12} \right).$
고도는 $\overline{AD},$ $\overline{BE},$ 및 $\overline{CF}.$로 합니다.

[asy]
unitsize(0.6 cm);

쌍 A, B, C, D, E, F, H;

B = (0,0);
C = (8,0);
A = 교차점(arc(B,5,0,180),arc(C,7,0,180));
H = 직교중심(A,B,C);
D = (A + reflect(B,C)*(A))/2;
E = (B + reflect(C,A)*(B))/2;
F = (C + 반사(A,B)*(C))/2;

draw(A--B--C--cycle);
draw(A--D,점선);
draw(B--E,점선);
draw(C--F,점선);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NE);
label("$F$", F, NW);
label("$H$", H, SE, UnFill);

dot(H);
[/asy]

코사인의 법칙에 따라,
\begin{align*}
\cos A &= \frac{5^2 + 7^2 - 8^2}{2 \cdot 5 \cdot 7} = \frac{1}{7}, \\.
\cos B &= \frac{5^2 + 8^2 - 7^2}{2 \cdot 5 \cdot 8} = \frac{1}{2}, \\
\cos C &= \frac{7^2 + 8^2 - 5^2}{2 \cdot 7 \cdot 8} = \frac{11}{14}.
\end{align*}그렇다면 $BD = AB \cos B = \frac{5}{2}$ 이고 $CD = AC \cos C = \frac{11}{2},$ 이므로 다음과 같습니다.
\[\overrightarrow{D} = \frac{11}{16} \오른쪽대각선{B} + \frac{5}{16} \또한 $AE = AB \cos A = \frac{5}{7}$ 및 $CE = BC \cos C = \frac{44}{7},$이므로 다음과 같습니다.
\[\overrightarrow{E} = \frac{44}{49} \오른쪽대각선{A} + \frac{5}{49} \overrightarrow{C}.\]이 방정식에서 $\overrightarrow{C}$ 를 분리하면 다음과 같이 구할 수 있습니다.
\[\overrightarrow{C} = \frac{16 \overrightarrow{D} - 11 \overrightarrow{B}}{5} = \frac{49 \overrightarrow{E} - 44 \overrightarrow{A}}{5}.\]그러면 $16 \overrightarrow{D} - 11 \overrightarrow{B} = 49 \overrightarrow{E}입니다. - 44 \overrightarrow{A},$이므로 $16 \overrightarrow{D} + 44 \overrightarrow{A} = 49 \overrightarrow{E} + 11 \overrightarrow{B},$ 또는
\[\frac{16}{60} \overrightarrow{D} + \frac{44}{60} \overrightarrow{A} = \frac{49}{60} \오른쪽대각선{E} + \frac{11}{60} \overrightarrow{B}.\]방정식의 양쪽 계수가 모두 1이 되므로 왼쪽의 벡터는 $AD,$ 라인에 있고 오른쪽의 벡터는 $BE,$ 라인에 있습니다. 따라서 이 공통 벡터는 $\overrightarrow{H}.$ 그러면 다음과 같습니다.
\begin{align*}
\overrightarrow{H} &= \frac{49}{60} \overrightarrow{E} + \frac{11}{60} \overrightarrow{B} \\
&= \frac{49}{60} \왼쪽( \frac{44}{49} \overrightarrow{A} + \frac{5}{49} \overrightarrow{C} \right) + \frac{11}{60} \overrightarrow{B} \\
&= \frac{11}{15} \overrightarrow{A} + \frac{11}{60} \overrightarrow{B} + \frac{1}{12} \overrightarrow{C}.
\end{align*}따라서 $(x,y,z) = \left( \frac{11}{15}, \frac{11}{60}, \frac{1}{12} \right).$입니다.