3. In the plane of the right triangle $\triangle A B C, \mathrm{~m}(\angle A)=90^{\circ}$, perpendiculars $A M$ and $B N$ are raised on the same side. Given that $A B=2 \sqrt{3} \mathrm{~cm}, A C=4 \mathrm{~cm}, A M=2 \mathrm{~cm}$ and $B N=1 \mathrm{~cm}$, determine the distance from point $M$ to the line of intersection of the planes $(A B C)$ and (MNC).
3. If $\{D\}=A B \cap M N$, then $(A B C) \cap(M N C)=C D$. From $\triangle A M D \sim \triangle B N D$ we obtain that $A D=4 \sqrt{3} \mathrm{~cm}$, so $C D=8 \mathrm{~cm}$. Let $A T$ be the altitude of the right triangle $A C D$; we have that $A T=\frac{A C \cdot A D}{C D}=2 \sqrt{3} \mathrm{~cm}$. According to the theorem of the three perpendiculars, $M T \perp C D$ and, from the right triangle $A M T$, we find that the distance from $M$ to the line $C D$ is $4 \mathrm{~cm}$.