3. For a regular tetrahedron $ABCD$ with edge length $1$, $E$ is a point inside $\triangle ABC$. The sum of the distances from point $E$ to the sides $AB$, $BC$, and $CA$ is $x$, and the sum of the distances from point $E$ to the planes $DAB$, $DBC$, and $DCA$ is $y$. Then $x^{2}+y^{2}$ equals ( ).
(A) 1
(B) $\frac{\sqrt{6}}{2}$
(C) $\frac{5}{3}$
(D) $\frac{17}{12}$
3.D.

Point $E$ to the distances of sides $A B, B C, C A$ sum up to the height of $\triangle A B C$, which is $\frac{\sqrt{3}}{2}$, hence $x=\frac{\sqrt{3}}{2}$.
$$
\begin{array}{l}
\frac{1}{3} S_{\triangle A B C} \cdot h \\
=\frac{1}{3} S_{\triangle \operatorname{MAB}} \cdot y_{1}+\frac{1}{3} S_{\triangle H C C} \cdot y_{2}+\frac{1}{3} S_{\triangle N C A} \cdot y_{3} \text {, } \\
\end{array}
$$

Here $h$ is the height of the regular tetrahedron, and $y_{1}, y_{2}, y_{3}$ are the distances from point $E$ to the planes $D A B$, $D B C$, $D C A$ respectively. Thus, $h=\frac{\sqrt{6}}{3}$. Therefore, $y_{1}+y_{2}+y_{3}=y$. So, $y=\frac{\sqrt{6}}{3}$.