I think that you should be checking:! \[ \begin{array}{c}\ \stackrel{\longrightarrow }{MA}\text{vol}(MBCD)+\stackrel{\longrightarrow }{MB}\text{vol}(MACD)+\stackrel{\longrightarrow }{MC}\text{vol}(MABD)+\stackrel{\longrightarrow }{MD}\text{vol}(MABC) =\boxed{ \overrightarrow{0}}\end{array} \] - Because: L.H.S is a vector while R.H.S is a number. We always have vector $\neq $ number! My solution: $\overrightarrow{MA}.V_{M.BCD}+\overrightarrow{MB}.V_{M.ACD}+\overrightarrow{MC}.V_{M.ABD}+\overrightarrow{MD}.V_{M.ABC}=\overrightarrow{0}$ $\Leftrightarrow$ $ \overrightarrow{MA}.\left (\frac{V_{M.BCD}}{V_{A.BCD}} \right )+\overrightarrow{MB}.\left (\frac{V_{M.ACD}}{V_{B.ACD}} \right )+\overrightarrow{MC}.\left (\frac{V_{M.ABD}}{V_{C.ABD}} \right ) + \overrightarrow{MD}.\left (\frac{V_{M.ABC}}{V_{D.ABC}} \right )=\overrightarrow{0}$ $\Leftrightarrow \overrightarrow{MA}.k_a+\overrightarrow{MB}.k_b+\overrightarrow{MC}.k_c+\overrightarrow{MD}.k_d=\overrightarrow{0}$; (where $k_a+k_b+k_c+k_d=1$) + Because: $ \overrightarrow{MA};\overrightarrow{MB};\overrightarrow{MC} \notin plane -(\alpha ).$ According to the theorem: "We denote a vector according three vectors are not in a plane" $\Rightarrow$ always $\exists (m,n,p): $ $\overrightarrow{MD}=m\overrightarrow{MA}+n\overrightarrow{MB}+p\overrightarrow{MC}$. We choice: $k_a=\frac{h_1}{h_a};k_b=\frac{h_2}{h_b};k_c=\frac{h_3}{h_c};k_d=\frac{h_4}{h_d};$ with: ($h_1,h_a$:The height of the tetrahedron $M.BCD$ and $A.BCD$, similar to $h_2;h_3;h_4;h_b;h_c;h_d$ ); Therefore: + $m=-\frac{k_a}{k_d}=-\frac{h_1.h_d}{h_a.h_4}=const;$ + $n=-\frac{k_b}{k_d}= -\frac{h_2.h_d}{h_b.h_4}=const;$ + $p=-\frac{k_c}{k_d} =-\frac{h_3.h_d}{h_c.h_4}=const;$ *) Now, we will prove only $\exists (m, n, p):$ $\overrightarrow{MA}.k_1+\overrightarrow{MB}.k_2+\overrightarrow{MC}.k_3=\overrightarrow{MD}; (k_1 \neq m;k_2 \neq n;k_3 \neq p) $ $\Rightarrow (k_1-m)\overrightarrow{MA}+(k_2-n)\overrightarrow{MB}+(k_3-p)\overrightarrow{MC}=\overrightarrow{0}\Rightarrow \overrightarrow{MA},\overrightarrow{MB} $ and $\overrightarrow{MC} \in plane-(ABC);$ (absurd). Because:"$M$ is interor point of $ABCD$". Therefore, $m=k_1;n=k_2;p=k_3$.

Denote $\vec{a}:=\overrightarrow{MA}$, $\vec{b}:=\overrightarrow{MB}$, $\vec{c}:=\overrightarrow{MC}$, and $\vec{d}:=\overrightarrow{MD}$. Because the four vectors belong in a three-dimensional vector space, they are not linearly independence. WLOG, suppose $\vec{d} \in \mathrm{span}\left\{\vec{a},\vec{b},\vec{c}\right\}$. It suffices to show the following identity: \[LHS:= \vec{a}\Big(\vec{b} \cdot \big(\vec{c} \times \vec{d}\big)\Big)-\vec{b}\Big(\vec{c} \cdot \big(\vec{d} \times \vec{a}\big)\Big)+\vec{c}\Big(\vec{d} \cdot \big(\vec{a} \times \vec{b}\big)\Big)-\vec{d}\Big(\vec{a} \cdot \big(\vec{b} \times \vec{c}\big)\Big)=0\,,\] whenever $\vec{d} \in \left\{\vec{a},\vec{b},\vec{c}\right\}$. However, this is quite easy. When $\vec{d}=\vec{a}$, we obtain \[LHS=\vec{a}\left(\vec{b}\cdot\big(\vec{c}\times\vec{a}\big) - \vec{a}\cdot\big(\vec{b}\times\vec{c}\big)\right)=0\,.\] Similarly, the claim is true for both $\vec{d}=\vec{b}$ and $\vec{d}=\vec{c}$. PS: Why do I have alternating signs for $LHS$ terms?

Dear amparvardi! This is a problem and interesting, I enjoyed this task, on our forum a bit less on geometry problems - $3D$ - space with flat geometry. You can post your solution! Your answer can make myself is learn more.