Prove that $XX’,YY’$ and $ZZ’$ are concurrent at $T$ 
Let $\Delta ABC$ has median lines respectively be $AX,BY,CZ$. And $X',Y',Z'$ be respectively midpoint of bisectors $AA', BB', CC'$. Prove that $XX',YY'$ and $ZZ'$ are concurrent and assume that point is $T$ and prove that $T$ is in the straight line connecting the center of the inscribed circle and the Lemoine point of the triangle $\Delta ABC$.



My teacher said that we could use Vector to solve it but u habeco no idea about that approach. Help me.

I will use barycentric coordinates of points, considered w.r.t. the given triangle $\Delta ABC$ with sides denoted by $a,b,c$. A point $P$ has then barycentric coordinates $(x,y,z)$, $x+y+z=1$, iff we have a relation of the form
$$
P = xA+yB +zC\ .
$$
(Using point affixes. This translates as the relation
$OP=xOA+yOB+zOC$ in terms of the vectors $OP;OA,OB,OC$ considered w.r.t. one (and any) fixed reference point $O$.)
Sometimes we use the notation $P(x:y:z)$ instead of $P(x,y,z)$, in this case the sum $x+y+z$ may or may not be normed to $1$, so this point is then explicitly $P(\ x/(x+y+z),\ y/(x+y+z),\ z/(x+y+z)\ )$.
A good reference is
Max Schindler, Evan Chen, Barycentric Coordinates for the Impatient.
Note that the point $T$ is in the ETC
the point

X(37) = CROSSPOINT OF INCENTER AND CENTROID

and that some lines after this information it claims

X(37) lies on these lines:
1,6   2,75   3,975   7,241 ...

and so on. Being on the line through 1, 6 (i.e. through the incenter 1 = X(1), and the symmedian = Lemoine point 6 = X(6) ) is exactly our problem. On this line there are also some other remarcable triangle centers, for instance X(9)...
Let us compute now...

The mid points of the sides have homogeneous coordinates
$$
\begin{aligned}
X&(0:1:1)\ , \\
Y&(1:0:1)\ , \\
Z&(1:1:0)\ .
\end{aligned}
$$
For instance, $X$ is then $\left(0,\frac12,\frac12\right)$ after passing to normed coordinates, corresponding to
$X=0A+\frac 12B+\frac 12C=\frac 12(B+C)$.
The angle bisectors of the angles $A,B,C$ intersect the opposite sides in $A',B',C'$ with coordinates
$$
\begin{aligned}
A'&(0:b:c)\ , \\
B'&(a:0:c)\ , \\
C'&(a:b:0)\ .
\end{aligned}
$$
For instance, $A'=\frac b{b+c}B+\frac c{b+c}C$, thus a point on $BC$, and the ratio $BA':A'C=c:b$, as stipulated by the angle bisector theorem.
The intersection $I$ of the angle bisectors is $(a:b:c)$, because it is on each line $AA'$, $BB'$, $CC'$, because the determinant corresponding to the barycentric coordinates of the points $A$, $A'$, and $I$ (presumably $(a:b:c)$),
$$
\begin{vmatrix}
1 &0&0\\0&b&c\\a&b&c
\end{vmatrix}
=0
$$
vanishes, and the same happens by sinymmetry also for the triples $A,B',I$, and $C,C',I$.
The mid points of the angle bisectors are now
$$
\begin{aligned}
X' &=\frac 12(A+A')\\
&=\frac 12(1,0,0)+\frac 12\cdot\frac 1{b+c}(0,b,c)\\
&=\frac 1{2(b+c)}(b+c,b,c)\\
&= (b+c:b:c)\ , \\
Y'&=(a:a+c:c)\ ,\\
Z'&=(a:b:a+b)\ .
\end{aligned}
$$
The point of intersection of three lines $XX',YY',ZZ'$ is, after solving a system of linear equation (as in the check in the sequel):
$$
T = \Big(\ a(b+c),\ b(a+c),\ c(a+b)\ \Big)\ .
$$
To see that $T$ is on $XX'$ we compute the determinant with rows given by the barycentric coordinates of the three points,
$$
\begin{aligned}
\begin{vmatrix}
0 & 1 & 1\\
b+c & b & c\\
a(b+c) & b(a+c)& c(a+b)
\end{vmatrix}
&=
(b+c)
\begin{vmatrix}
0 & 1 & 1\\
1 & b & c\\
a & b(a+c)& c(a+b)
\end{vmatrix}
\\
&=
(b+c)
\begin{vmatrix}
0 & 1 & 1\\
1 & b & c\\
0 & b(a+c)-ab& c(a+b)-ac
\end{vmatrix}
\\
&=
bc(b+c)
\begin{vmatrix}
0 & 1 & 1\\
1 & b & c\\
0 &  1& 1
\end{vmatrix}
\\
&=0\ .
\end{aligned}
$$
By (cyclic) symmetry, the point $T$ also lies on $YY'$ and $ZZ'$.

We have to show finally that $T=X(37)$, as labeled in the Encyclopedia of Triangle Centers, ETC, is on the line through the incenter $X(1)=I(a:b:c)$, and the symmedian = Lemoine point $X(6)=L=(a^2:b^2:c^2)$, so we compute one more easy determinant, built using the barycentric coordinates of the points $I, L, T$:
$$
\begin{aligned}
\begin{vmatrix}
a & b & c\\
a^2 & b^2 & c^2 \\
a(b+c) & b(a+c)& c(a+b)
\end{vmatrix}
&=
abc
\begin{vmatrix}
1 & 1 & 1\\
a & b & c\\
b+c & a+c & a+b
\end{vmatrix}
\\
&=
abc
\begin{vmatrix}
1 & 1 & 1\\
a & b & c\\
a+b+c & a+b+c & a+b+c
\end{vmatrix}
\\
&=
abc(a+b+c)
\begin{vmatrix}
1 & 1 & 1\\
a & b & c\\
1 & 1 & 1
\end{vmatrix}
\\
&=0\ .
\end{aligned}
$$

Note: This is an analytic solution. (A solution that is in the same time algebraic, barycentric, vectorial, analytic.. These attributes are in this case hard to distinguish, we are performing "blind geometry" without any (need of a) picture.)
If a synthetic solution is also wanted, i can try to type and insert it.
