A) \[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\]
B) \[x+y+z=1\]
C) \[\frac{1}{1-x}+\frac{1}{1-y}+\frac{1}{1-z}=1\]
D) None of these
Correct Answer: A
Solution :
| \[A(2-x,2,2),\,\,B(2,2-y,2),\,\,C(2,2,-z),\,\,D(1,1,1)\] |
| \[\overrightarrow{DA}=(1-x)\hat{i}+\hat{j}+\hat{k}\,\,;\,\,\overrightarrow{DB}=\hat{i}+(1-y)\hat{j}+\hat{k}\,\,;\] |
| \[\,\overrightarrow{DC}=\hat{i}+\hat{j}+(1-z)\hat{k}\] |
| If four points are coplanar then |
 \[[\overrightarrow{DA},\overrightarrow{DB},\overrightarrow{DC}]=0\] \[\Rightarrow \left| \begin{matrix} 1-x & 1 & 1 \\ 1 & 1-y & 1 \\ 1 & 1 & 1-z \\ \end{matrix} \right|=0\] |
| \[{{c}_{1}}\to {{c}_{1}}-{{c}_{2}}\]and\[{{c}_{2}}\to {{c}_{2}}-{{c}_{3}}\]\[\Rightarrow \left| \begin{matrix} -x & 0 & 1 \\ y & -y & 1 \\ 0 & z & 1-z \\ \end{matrix} \right|=0\] |
| \[\therefore \,\,\,-x(-y+yz-z)+1(+yz)=0\,\,;\] |
| \[xy-xyz+xz+yz=0\] |
| \[xy+yz+zx=xyz\] |
| \[\therefore \,\,\,\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\] |
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