We have a parallelogram $ABCD$,

whose vertices are $A\left(2,4,-1\right), B\left(3,6,-1\right), C\left(4,5,1\right)$

Now, assume the fourth vertex $D\left(x,y,z\right)$

So, we have diagonals $AC$ and $BD$

Now, we know that, diagonals of a parallelogram bisect each other,

So, coordinate of $E$, as mid-point of $AC$ is $\left(3, \frac{9}{2}, 0\right) ...\left(1\right)$

As mid-point of $BD$ is $\left(\frac{x+3}{2}, \frac{y+6}{2}, \frac{z-1}{2}\right) ...\left(2\right)$
So, comparing $\left(1\right)$ and $\left(2\right)$, 

$\frac{x+3}{2}=3 \Rightarrow x=3$

$\frac{y+6}{2}=\frac{9}{2}\Rightarrow y=3$ and 

$\frac{z-1}{2}=0 \Rightarrow z=1$

So, the coordinate of $D$ is $\left(3, 3, 1\right)$.