If three consecutive vertices of parallelogram are (3.4.-1) ,(7,10,-3) and (8,1,0),then find the coordinate of fourth vertex..dont giveany links

The three vertices of a parallelogram ABCD are given as A (3, 4, -1), B(7,10,-3), and C (8,1,0). Let the coordinates of the fourth vertex be D (x, y, z).

We know that the diagonals of a parallelogram bisect each other.

Therefore,in parallelogram ABCD, AC and BD bisect each other.

∴Mid-point of AC = Mid-point of BD

$$\begin{gathered} \Rightarrow \left(\frac{3+8}{2},\frac{4+1}{2},\frac{-1+0}{2}\right)=\left(\frac{\mathrm{x}+7}{2},\frac{\mathrm{y}+10}{2},\frac{\mathrm{z}-3}{2}\right) \\ \Rightarrow \left(\frac{11}{2},\frac{5}{2},\frac{-1}{2}\right)=\left(\frac{\mathrm{x}+7}{2},\frac{\mathrm{y}+10}{2},\frac{\mathrm{z}-3}{2}\right) \\ \mathrm{Now} \mathrm{comparing} \mathrm{the} \mathrm{coordinates} \mathrm{we} \mathrm{get} \\ \frac{\mathrm{x}+7}{2}=\frac{11}{2} \\ \Rightarrow \mathrm{x}=4 \\ \mathrm{and} \frac{\mathrm{y}+10}{2}=\frac{5}{2} \\ \Rightarrow \mathrm{y}=-5 \\ \mathrm{and} \frac{\mathrm{z}-3}{2}=\frac{-1}{2} \\ \Rightarrow \mathrm{z}=2 \end{gathered}$$