One, two, three, let's do it

Using the formula $\boxed{\text{Area}=\frac{1}{2} ab\sin C}$ we get:

$\therefore$ the area of $\triangle DEF$ is $1-\frac{17}{24}=\frac{7}{24}$ from $\triangle ABC$ so $\triangle DEF$ area is $\boxed{7}$

We know that area of triangle is directly proportional to its base and height.

Since $\triangle BDF$ has a base $\frac{2}{3}$ and height $\frac{1}{4}$ of those of $\triangle ABC$ , the area of $\triangle BDF = \frac{2}{3} \times \frac{1}{4} \times 24 = 4$ .

Similarly, the area of $\triangle ADE = \frac{1}{3} \times \frac{1}{2} \times 24 = 4$ .

And, the area of $\triangle CEF = \frac{1}{2} \times \frac{3}{4} \times 24 = 9$ .

Therefore, the area of $\triangle DEF = 24-4-4-9 = \boxed{7}$ .

My solution uses some heavy machinery to crack a not very tough nut. I hope you like it!

I will use pairs of letters to stand for vectors in the obvious way. Then the required area in terms of the vector cross product is

$\triangle DEF=\dfrac{1}{2}\left|DE\times DF\right|$

$\implies \triangle DEF=\dfrac{1}{2}\left|(\dfrac{1}{3}BA+\dfrac{1}{2}AC)\times (\dfrac{2}{3}AB+\dfrac{1}{4}BC)\right|$

now remembering that the vector product of parallel vectors is zero, and recognising that $AC\times BC=CA\times CB$

$\triangle DEF=\dfrac{1}{2}\left|\dfrac{1}{12}BA\times BC+\dfrac{1}{3}AC\times AB+\dfrac{1}{8}CA\times CB\right|$

Now the cross products are all perpendicular to the plane of the paper, and the magnitude of each of the cross products is twice the area of the original triangle. But by the right hand rule you can see that the last term points in the opposite direction to the first two. So the above expression becomes

$=\dfrac{2\times 24}{2}\left|\dfrac{1}{12}+\dfrac{1}{3}-\dfrac{1}{8}\right|$

$=|2+8-3|=\boxed{7}$

There exists an area-preserving linear transformation $T$ that maps the triangle ABC into the triangle (0,0), (6,0), and (0,8), with area 24, and the triangle DEF into (3,4), (0,2), and (4,0), with area $\boxed{7}$

Let $\theta = \angle CAB, \beta = \angle ACB$ and $\phi = \angle ABC$ . We have

$24=\dfrac{1}{2}(3y)(2x)(\sin \theta)$ $\implies$ $xy\sin \theta = 8$

$A[AED]=\dfrac{1}{2}xy\sin \theta = \dfrac{1}{2}(8)=4$

$24=\dfrac{1}{2}(2x)(4z)(\sin \beta)$ $\implies$ $xz\sin \beta = 6$

$A[ECF]=\dfrac{1}{2}(x)(3z)(\sin \beta)=\dfrac{3}{2}(6)=9$

$24=\dfrac{1}{2}(3y)(4z)(\sin \phi)$ $\implies$ $yz\sin \phi=4$

$A[FBD]=\dfrac{1}{2}(z)(2y)(\sin \phi)=zy\sin \phi = 4$

Therefore,

$A[DEF]=24-A[AED]-A[ECF]-A[FBD]=24-4-9-4=\boxed{7}$

Take a special case: use right triangle with sides $6,8,10$ whose area is $24$

Configure the lengths of the divided segments. This gives 4 triangles: 1 smaller right triangle, and 3 scalene triangles including the one we're looking for.

Draw the altitudes of the 2 scalene triangle so (excluding the one we're asked to find). This forms even smaller right triangles inside the scalene triangles. Use right triangle proportions to find the lengths of the altitudes. Find the area of the 2 scalene triangles and the small right triangle, add them up and subtract the sum from 24 and you'll get $\boxed{7}$

Add a dot G in the middle of C and B. Dot G will be $2$ z from C and D, and line EG will be parallel to AB. AC is twice the length of CE, so the area of ABC is $4$ times of CEG, so we know that CEG is $6$ and AEGB will be $18$ . EG will be half of AB, which is $3$ y / $2$ . Assume the distance between EG and AB to be d , area AEGB will be the area sum of triangle AEB and EGB $3$ yd / $2$ + $3$ yd / $4$ = $18$ . we can find that yd / $2=4$ , AED and DFB will be $4$ ,EGF will be $3$ . Subtract all and EFD will be \boxed{7}.

peasant's solution:
1. Make the given triangle an isosceles right triangle with vertices at (0,0) (0,4sqrt3) and (4sqrt3,0)
2. Using analytic geometry, the vertices of triangle DEF are (0,2sqrt3) (4qrt3 / 3, 0) and (3sqrt3, sqrt3)
3. Then use the formula for solving area of triangle given three vertices (determinant method)