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Soln 0^{-}\begin{array}{l}\left(x_{1}, y_{1}, z, 1\right)=(0,3,5) \\(x, y, y, z, z)=(7,1,3) \\D(7,1,3) \begin{array}{l}A, 3,5) \\\left(x_{2}, y_{2}, 2,2\right)=(-3,-1,17) \\C\left(x_{3}, y_{3}, 23\right)=(4,-3,15)\end{array} \\A B=d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\=\sqrt{(-3-0)^{2}+(-1-3)^{2}+(17-5)^{2}} \\=\sqrt{9+16+144} \text { व }=\sqrt{169}\end{array}\therefore A B=13\begin{aligned}B C & =\sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}+\left(z_{3}-z_{2}\right)^{2}} \\& =\sqrt{(4+3)^{2}+(-3+1)^{2}+(15-17)^{2}} \\& \Rightarrow \sqrt{49+4+4}=\sqrt{57}\end{aligned}\therefore \quad B C=\sqrt{57}\begin{aligned}C D & =\sqrt{\left(x_{4}-x_{3}\right)^{2}+\left(y 4-y_{3}\right)^{2}+\left(z_{4}-z_{3}\right)^{2}} \\& =\sqrt{(7-4)^{2}+(1+3)^{2}+(3-15)^{2}} \\& =\sqrt{9+16+144}=\sqrt{169}=13 \\\therefore C D & =13\end{aligned}\begin{aligned}D A & =\sqrt{\left(x_{1}-x_{4}\right)^{2}+\left(y_{1}-y_{4}\right)^{2}+\left(z_{1}-z_{4}\right)^{2}} \\& \Rightarrow \sqrt{(0-7)^{2}+(3-1)^{2}+(5-3)^{2}} \\& \Rightarrow \sqrt{49+4+4} \Rightarrow \sqrt{57}\end{aligned}Thus, oppasite sides of quadrilateral are equalA B=C D \text { \& } B C=D AWe know that, in case of parallelogram appasite sides are equal & diagonals bisects each otheer.\text { midpoint of } \begin{aligned}A C & =\left(\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}, \frac{z_{1}+z_{3}}{2}\right) \\& =\left(\frac{0+4}{2}, \frac{3-3}{2}, \frac{5+15}{2}\right) \\& =(2,0,10)\end{aligned}\text { Midpoint of } \begin{aligned}B D & =\left(\frac{x_{2}+x_{4}}{2}, \frac{y_{2}+y_{4}}{2}, \frac{z_{2}+z_{4}}{2}\right) \\& =\left(\frac{-3+7}{2}, \frac{-1+1}{2}, \frac{17+3}{2}\right) \\& =(2,0,10)\end{aligned}midpoint of A C= midpoint of B D\therefore oppasite sides are equel & diagonal bisect each othee. So we can say that A B C D is a parall elogram ...