In the tetrahedron $S-ABC$, $SA \perp$ plane $ABC$, $AB=BC=SA=1$, $\angle ABC=\frac{\pi}{2}$.
(1) Find the sine value of the angle between $SC$ and plane $ABC$;
(2) Find the distance from $B$ to plane $ASC$;
(3) Find the size of the acute dihedral angle between planes $SBC$ and $SAC$.


18 On the plane $ABC$, draw $AD \perp AB$. Taking $A$ as the origin, and the directions of vectors $AB$, $AD$, and $AS$ as the $x$, $y$, and $z$ axes respectively, we establish a spatial rectangular coordinate system. Thus, we have $A(0,0,0)$, $B(1,0,0)$, $C(1,1,0)$, $S(0,0,1)$.
(1) Since $SA \perp$ plane $ABC$, the angle between $SC$ and plane $ABC$ is $\angle ACS$. In the right $\triangle SAC$, $AC=\sqrt{2}$, $SC=\sqrt{3}$, so
$$
\sin \angle ACS=\frac{SA}{SC}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}.
$$
(2) Let the normal vector of plane $ASC$ be $\vec{n}=(x, y, z)$, then $\vec{n} \perp \overrightarrow{AS}$ and $\vec{n} \perp \overrightarrow{SC}$, and $\overrightarrow{AS}=(0,0,1)$, $\overrightarrow{SC}=(1,1,-1)$, so $\left\{\begin{array}{l}z=0, \\ x+y-z=0,\end{array}\right.$ thus $\vec{n}=(1,-1,0)$. Therefore, the distance from $B$ to plane $ASC$ is
$$
d=\frac{|\overrightarrow{AB} \cdot \vec{n}|}{|\vec{n}|}=\frac{\sqrt{2}}{2}.
$$
(3) Let the normal vector of plane $SBC$ be $\vec{m}=(x, y, z)$, then $\vec{m} \perp \overrightarrow{BC}$ and $\vec{m} \perp \overrightarrow{SC}$, and $\overrightarrow{BC}=(0,1,0)$, so $\left\{\begin{array}{l}y=0, \\ x+y-z=0,\end{array}\right.$ thus $\vec{m}=(1,0,1)$. Let the acute dihedral angle between plane $SBC$ and $SAC$ be $\theta$, then
$$
\cos \theta=\frac{|\vec{m} \cdot \vec{n}|}{|\vec{m}| \cdot|\vec{n}|}=\frac{1}{2},
$$

i.e.,
$$
\theta=\frac{\pi}{3}.
$$