We see that $\mathcal{T}$ is the triangle whose vertices are $(1,0,0),$ $(0,1,0),$ and $(0,0,1).$  We are looking for the points $(x,y,z) \in \mathcal{T}$ such that exactly two of the following inequalities hold: $x \ge \frac{1}{2},$ $y \ge \frac{1}{3},$ and $z \ge \frac{1}{6}.$

The plane $x = \frac{1}{2}$ cuts triangle $\mathcal{T}$ in a line that is parallel to one of its sides.  The same holds for the planes $y = \frac{1}{3}$ and $z = \frac{1}{6}.$

Let $\mathcal{A}$ be the set of points in $\mathcal{T}$ such that $x \ge \frac{1}{2}$ and $y \ge \frac{1}{3}.$  Then the inequality $z \le \frac{1}{6}$ is automatically satisfied, and $z = \frac{1}{6}$ only for the point $\left( \frac{1}{2}, \frac{1}{3}, \frac{1}{6} \right).$  Thus, $\mathcal{A}$ is a triangle which is similar to $\mathcal{T},$ and the ratio of their areas is $\frac{1}{6^2} = \frac{1}{36}.$

[asy]
import three;

size(220);
currentprojection = perspective(6,3,2);

triple P = (1/2,1/3,1/6), Q = (5/6,0,1/6), R = (1/2,0,1/2), S = (0,1/3,2/3), T = (0,5/6,1/6), U = (1/2,1/2,0), V = (2/3,1/3,0);

draw(surface(P--Q--R--cycle),paleyellow,nolight);
draw(surface(P--S--T--cycle),paleyellow,nolight);
draw(surface(P--U--V--cycle),paleyellow,nolight);
draw((1,0,0)--(0,1,0)--(0,0,1)--cycle);
draw((0,0,0)--(1,0,0),dashed);
draw((0,0,0)--(0,1,0),dashed);
draw((0,0,0)--(0,0,1),dashed);
draw(Q--T);
draw(R--U);
draw(S--V);
draw((1,0,0)--(1.2,0,0),Arrow3(6));
draw((0,1,0)--(0,1.2,0),Arrow3(6));
draw((0,0,1)--(0,0,1.2),Arrow3(6));

label("$x$", (1.3,0,0));
label("$y$", (0,1.3,0));
label("$z$", (0,0,1.3));
label("$x = \frac{1}{2}$", R, W);
label("$y = \frac{1}{3}$", S, NE);
label("$z = \frac{1}{6}$", T, NE);
label("$\mathcal{A}$", (P + U + V)/3);
label("$\mathcal{B}$", (P + Q + R)/3);
label("$\mathcal{C}$", (P + S + T)/3);
[/asy]

Likewise, let $\mathcal{B}$ be the set of points in $\mathcal{T}$ such that $x \ge \frac{1}{2}$ and $z \ge \frac{1}{6},$ and let $\mathcal{C}$ be the set of points in $\mathcal{T}$ such that $y \ge \frac{1}{3}$ and $z \ge \frac{1}{6}.$  Then $\mathcal{B}$ and $\mathcal{C}$ are triangles that are also similar to $\mathcal{T},$ and the ratio of their areas to the area of $\mathcal{T}$ are $\frac{1}{3^2} = \frac{1}{9}$ and $\frac{1}{2^2} = \frac{1}{4},$ respectively.

Therefore, the area of $\mathcal{S}$ divided by the area of $\mathcal{T}$ is $\frac{1}{36} + \frac{1}{9} + \frac{1}{4} = \frac{7}{18}.$
우리는 $\mathcal{T}$가 꼭지점이 $(1,0,0),$ $(0,1,0),$ 및 $(0,0,1)인 삼각형이라는 것을 알 수 있습니다.다음 부등식 중 정확히 두 가지가 성립하는 $x \ge \frac{1}{2},$ $y \ge \frac{1}{3},$ 및 $z \ge \frac{1}{6}.$ 점을 \mathcal{T}$에서 찾고 있습니다.

x = \frac{1}{2}$ 평면은 삼각형 $\mathcal{T}$를 그 변 중 하나에 평행한 선으로 잘라냅니다.  평면 $y = \frac{1}{3}$와 $z = \frac{1}{6}.$도 마찬가지입니다.

x \ge \frac{1}{2}$, $y \ge \frac{1}{3}$가 되는 $\mathcal{T}$의 점 집합을 $\mathcal{A}$라고 합니다.그러면 부등식 $z \le \frac{1}{6}$는 자동으로 만족되며, $z = \frac{1}{6}$는 $\left( \frac{1}{2}, \frac{1}{3}, \frac{1}{6} \right)의 점에 대해서만 충족됩니다.$ 따라서 $\mathcal{A}$는 $\mathcal{T}$와 유사한 삼각형이며, 그 면적의 비율은 $\frac{1}{6^2} = \frac{1}{36}.$입니다.

[asy]
세 개를 가져옵니다;

size(220);
currentprojection = perspective(6,3,2);

triple P = (1/2,1/3,1/6), Q = (5/6,0,1/6), R = (1/2,0,1/2), S = (0,1/3,2/3), T = (0,5/6,1/6), U = (1/2,1/2,0), V = (2/3,1/3,0);

draw(surface(P--Q--R--cycle),paleyellow,nolight);
draw(surface(P--S--T--cycle),paleyellow,nolight);
draw(surface(P--U--V--cycle),paleyellow,nolight);
draw((1,0,0)--(0,1,0)--(0,0,1)--cycle);
draw((0,0,0)--(1,0,0),dashed);
draw((0,0,0)--(0,1,0),점선);
draw((0,0,0)--(0,0,1),점선);
draw(Q--T);
draw(R--U);
draw(S--V);
draw((1,0,0)--(1.2,0,0),Arrow3(6));
draw((0,1,0)--(0,1.2,0),Arrow3(6));
draw((0,0,1)--(0,0,1.2),Arrow3(6));

label("$x$", (1.3,0,0));
label("$y$", (0,1.3,0));
label("$z$", (0,0,1.3));
label("$x = \frac{1}{2}$", R, W);
label("$y = \frac{1}{3}$", S, NE);
label("$z = \frac{1}{6}$", T, NE);
label("$\mathcal{A}$", (P + U + V)/3);
label("$\mathcal{B}$", (P + Q + R)/3);
label("$\mathcal{C}$", (P + S + T)/3);
[/asy]

마찬가지로, $\mathcal{B}$ 를 $\mathcal{T}$ 의 점 집합으로 $x \ge \frac{1}{2}$ 와 $z \ge \frac{1}{6},$ 라고 하고, $\mathcal{C}$ 를 $\mathcal{T}$ 의 점 집합으로 $y \ge \frac{1}{3}$ 와 $z \ge \frac{1}{6}$ 라고 합시다.그러면 $\mathcal{B}$ 와 $\mathcal{C}$ 는 $\mathcal{T}$ 와 유사한 삼각형이며, $\mathcal{T}$ 의 넓이에 대한 이들의 넓이의 비율은 각각 $\frac{1}{3^2} = \frac{1}{9}$ 와 $\frac{1}{2^2} = \frac{1}{4},$ 입니다.

따라서 $\mathcal{S}$의 넓이를 $\mathcal{T}$의 넓이로 나눈 값은 $\frac{1}{36} + \frac{1}{9} + \frac{1}{4} = \frac{7}{18}.$입니다.