11. 6. Can the incenter of the four faces of a tetrahedron be coplanar?
11.6. Not Coplanar.

Let $A, B, C, D$ be the four vertices of a tetrahedron, and $I_{A}$, $I_{B}$, $I_{C}$, $I_{D}$ be the incenters of $\triangle BCD$, $\triangle ACD$, $\triangle ABD$, and $\triangle ABC$, respectively. Suppose they are coplanar. Clearly, any three of them are not collinear, so they form the vertices of a convex quadrilateral or one of them lies inside the triangle formed by the other three.
(1) Without loss of generality, we can assume that the quadrilateral $I_{A} I_{B} I_{C} I_{D}$ is convex. Then the line segments $I_{A} I_{C}$ and $I_{B} I_{D}$ intersect. Let $M$, $N$, $K$, $L$ be the midpoints of edges $AB$, $AD$, $CD$, and $BC$, respectively. Draw a line $l$ through point $B$ parallel to $AC$. Then the incircle of $\triangle ABC$ lies between $l$ and $AC$, touching $AC$ but not touching $l$.

This indicates that points $I_{D}$ and $B$ are on opposite sides of the midline $ML$.

Similarly, points $I_{B}$ and $I_{D}$ are on one side of the plane $MNKL$, and points $I_{A}$ and $I_{C}$ are on the other side.
This indicates that the line segments $I_{A} I_{C}$ and $I_{B} I_{D}$ do not intersect, which is a contradiction.
(2) Point $I_{A}$ lies inside $\triangle I_{B} I_{C} I_{D}$. Note that points $I_{B}$, $I_{C}$, and $I_{D}$ are on one side of the plane $BCD$, while $I_{A}$ lies on this plane, so this situation cannot occur.