Here's another way (it's always good to know more than one way to solve a problem).
Your plane goes through $(-1,5,2)$ and $(2,7,3)$ (obtained by taking $t=0$ and $t=1$, respectively) and also $(2,4,-1)$. If your plane is $$ax+by+cz+d=0$$ then each of these three points gives you an equation relating the four unknowns $a,b,c,d$. So you get three homogeneous linear equations in 4 unknowns. Presumably, you know how to solve such a system. You'll get a one-parameter family of solutions (because $akx+bky+ckz+dk=0$ is the same plane, for any non-zero value of $k$); just pick any member of this family.
Set the two equations equal to each other. Then for each of the three coordinates, you will get an equation in $s$ and $t$, so you will have three equations in two variables. If the two lines are co-planer then there is a unique solution for $s$ and $t$ which will give you the coordinates of the point of intersection.
For example, the first coordinates give you the equation
$$ 2+3t=5-3s $$
Find the equations for the other two coordinates and finish the problem.
ADDENDUM:
Now that you correctly found the point of intersection $(5,0,3)$ you have the necessary information to find the equation of the plane which contains the two intersecting lines.
To find the equation of a plane containing two intersecting lines you need three pieces of information: direction vectors for each of the two lines and the point of intersection of the two lines.
The direction vectors are the vector coefficients of your two vector line equations:
- $\langle 3,-3,3\rangle$
- $\langle 3,-3,0\rangle$
These two may be simplified by multiplying by $\dfrac{1}{3}$ since multiplication by a nonzero constant does not change the direction of a vector. So use the following for the two direction vectors.
- $\langle 1,-1,1\rangle$
- $\langle 1,-1,0\rangle$
The cross-product of these two vectors gives a normal vector $N=\langle a,b,c\rangle$ for the plane containing the two lines.
$N=\begin{vmatrix}
\mathbf{i}&\phantom{-}\mathbf{j}&\mathbf{k}\\1&-1&1\\1&-1&0\end{vmatrix}=\mathbf{i}+\mathbf{j}=\langle 1,1,0\rangle$
The equation of a plane has the form
$$ ax+by+cz=d $$
where the normal vector is $\langle a,b,c\rangle$ and $d$ is computed using the values of $a,b,c$ and the coordinates of a point in the plane ( in this case, $(5,0,3)$).
So the equation for this plane is
$$ 1\cdot x+1\cdot y+0\cdot z=d$$
where
$$1(5)+1(0)+0(3)=d=5$$
So the equation of the line containing the two lines is
$$ x+y=5 $$
Best Answer
The line $\frac{x-2}{2} = \frac{y+4}{3} = \frac{2-z}{5}\,$ can be written as $$x=2+2t\,,\quad y=-4+3t \,,\quad z=2-5t \,, $$ and the line $$x = 3 + 4t \,,y = -4 + 6t \,,z = 5 - 10t \,.$$
The two lines have the same direction, since $ v_1=(2,3,-5) $ and $ v_2 = (4, 6,-10)\,, $ where $v_1$ and $v_2$ are the direction vectors of the two lines.
One can get two points lie in the plane. Putting $t=0$ in the equations of the lines gives $p_1=(2,-4,2)$ and $p_2=(3,-4,5)\,,$ which lie in the plane.
Constructing the vector $ v_3=p_2-p_1$ gives $v_3=(1,0,3)\,.$ Now, we can find the normal to the plane by taking the cross product of $v_3$ and $v_1$ or $v_2$.
$$ n = v_3 \times v_2 \,. $$
Once that done, the equation of the plane is given by
$$ n.(X-p_1)=0 \Rightarrow n.( x-2,y-4,z-2 )=0 \,.$$