Alright Tetrahedron

WLOG we assume A is at the origin, B is at (6,0,0) and C is at (0,10,0), so that the base ABC (of area 30) lies in the xy-plane. To get the volume of ABCD we just need the z-height of D. Clearly M is at (3,0,0), and if MN is orthogonal to AB we must have N = (3,x,y) for some x,y. Since N is the midpoint of CD this puts D at (6,2x-10,2y). Since $|MN|=4$ we have $x^2+y^2=16$ , while $|BD|$ =14 gives us $(x-5)^2+y^2=49$ . Subtracting these yields x = -4/5, and then y = 8√6/5. The height of D is therefore 2y=16√6/5, giving a volume of [ABCD] = 32√6.

We have that the volume of the tetrahedron $NABD$ is the half than the volume of the tetrahedron $ABCD$ because the base $ABD$ is in common and the height from $N$ is half than the height from $C$ because $ND = \frac{CD}{2}$ . The volume of the tetrahedron $CABN$ is equal to the volume of the tetrahedron $NABD$ that is half than the volume of $ABCD$ . By considering the base $ABC$ of the tetrahedron $ABCN$ we have that the height from $N$ falls on the axis of $AB$ because $MN$ is perpendicular to $AB$ .

We call $L$ the midpoint of $BC$ and we have that the axis of $AB$ meets $BC$ in $L$ because the triangle $MBL$ is similar to the triangle $ABC$ and the rate is $\frac{1}{2}$ . Now we consider the triangle $NML$ and we suppose $\angle NML=\alpha$ . We have that $NL=7$ because the triangle $CNL$ is similar to the triangle $CDB$ and the rate is $\frac{1}{2}$ and $ML=5$ for the same reason. So we have $NL^2=MN^2+ML^2-2MN\cdot ML\cos(\alpha)$ for the cosine rule. So $\cos(\alpha)=-\frac{1}{5}$ and $\sin(\alpha)=\frac{2\sqrt{6}}{5}$ .

Now we have that the height from $N$ to $ABC$ is given by $NM\sin(\alpha)$ and the area $S$ of $ABC$ is $30$ ( $AB$ and $AC$ are catheti). So the volume of $ABCN$ is given by $\frac{S\cdot NM\sin(\alpha)}{3}$ and so the volume of $ABCN$ is $16\sqrt{6}$ . So the volume of the tetrahedron $ABCD$ is $32\sqrt{6}$ . So $a=32$ and $b=6$ and so $a+b=38$ .

[Note: It is easier to calculate that $[MLN]= 4 \sqrt{6}$ , so the volume of $MLNB$ is $\frac {1}{3} \cdot |MB| \cdot [MLN] = 4 \sqrt{6}$ , hence the volume of $ABCN$ is 4 times that, which is $16 \sqrt{6}$ . - Calvin]

Suppose $L$ is the midpoint of $BC$ . Then $ML$ is parallel to $AC,$ so perpendicular to $AB.$ Because $MN$ is perpendicular to $AB,$ the plane $MLN$ is perpendicular to $AB$ . Note that the length of $ML$ is half of the length of $AC,$ so it is $5$ . Similarly, the length of $NL$ is $7$ , the triangle $MLN$ has side lengths $4,$ $5,$ and $7.$

Lemma. The area of the triangle $MLN$ is $4\sqrt{6}$ .

Proof of the Lemma. Suppose the measure of the angle $MLN$ is $\alpha$ . From the Law of Cosines, $\cos \alpha = \frac{4^2+5^2-7^2}{2\cdot 2\cdot 5}=-\frac{1}{5}$ . So $\sin \alpha = \sqrt{1-\cos ^2 \alpha} = \frac{\sqrt{24}}{5}$ . The area of $MLN$ is $\frac{1}{2}\cdot \frac{\sqrt{24}}{5} \cdot 4 \cdot 5 = 2\sqrt{24}=4\sqrt{6}$ .

Because $AB$ is perpendicular to $MLN,$ the segment $BM$ is the height of the tetrahedron $MLNB$ , when viewed with the base $MLN$ . So the volume of $MLNB$ is $\frac{1}{3} \cdot 3 \cdot 4\sqrt{6}=4\sqrt{6}$ .

Consider the tetrahedra $ABCN$ and $MBLN$ . They have the same vertex $N$ and their faces opposite to $N$ lie in the same plane. Because the triangle $ABC$ is similar to the triangle $MBL$ with the proportionality coefficient $2$ , the area of $ABC$ is four time the area of $MBL,$ so the volume of $ABNC$ is $4\cdot 4\sqrt{6}=16\sqrt{6}.$

Finally, consider the tetrahedra $ABCD$ and $ABCN$ as having the same vertex $A$ and opposite faces in the same plane. Clearly, the area of $BCD$ is twice the area of $BCN$ . So the volume of $ABCD$ is $2\cdot 16\sqrt{6}=32\sqrt{6},$ and the answer is $32+6=38.$

$V=\dfrac 1 3 *A_b*H.~~~~Area~ \Delta~ABC~=A_b=\dfrac 1 2 *6*10=30. \\Let ~ \Delta~ABC ~be~in~horizontal ~plane. \\\color{#D61F06}{\large \Delta~NML}\\MN\perp AB,~ so~ \Delta NML ~ is ~ in ~ the ~ vertical ~ plane.\\Let ~ h~be ~ the ~height ~ of ~ N ~ above ~ ML.\\ ML=\dfrac 1 2 *AC=5.~~~~~~~NL=\dfrac 1 2 *BD=7. ~~~~~~~ NM=4.\\\dfrac 1 2(5+7+4)=8.~~~~~~~~~\dfrac 1 2 *ML*h=its~area~by~Hero's.\\\implies~\dfrac 1 2 *5*h=\sqrt{8*(8-5)(8-7)(8-4) }=4*\sqrt 6 . \\h=\dfrac{2*4*\sqrt 6 }{5}.~~~~~~~~But ~ N ~ is ~ the ~ midpoint~ of~ CD .\\ \color{#D61F06}{\large V}\\\implies~H=2*h=\dfrac{16*\sqrt 6 }{5} ~~~~~~~~~~~V=\dfrac 1 3 *30* \dfrac{16*\sqrt 6 }{5} \\V=32*\sqrt 6 =a\sqrt b.~~~~~~~~~~~a+b =\color{#3D99F6}{\boxed {38} } \\~~\\\text{Thank you, Clvin Lin, it is very clear diagram }\\\text{making the problem solution easy. }$