If the points A(2-x, 2, 2),B(2, 2-y, 2),C(2, | vedclass

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(A) Four points $A, B, C, D$ are coplanar if the scalar triple product of vectors $\vec{DA}, \vec{DB}, \vec{DC}$ is zero. Given $A(2-x, 2, 2)$,$B(2, 2

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$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$

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(A) Four points $A, B, C, D$ are coplanar if the scalar triple product of vectors $\vec{DA}, \vec{DB}, \vec{DC}$ is zero. Given $A(2-x, 2, 2)$,$B(2, 2-y, 2)$,$C(2, 2, 2-z)$,and $D(1, 1, 1)$. $\vec{DA} = (2-x-1, 2-1, 2-1) = (1-x, 1, 1)$. $\vec{DB} = (2-1, 2-y-1, 2-1) = (1, 1-y, 1)$. $\vec{DC} = (2-1, 2-1, 2-z-1) = (1, 1, 1-z)$. The condition for coplanarity is $\begin{vmatrix} 1-x & 1 & 1 \ 1 & 1-y & 1 \ 1 & 1 & 1-z \end{vmatrix} = 0$. Expanding the determinant: $(1-x)[(1-y)(1-z) - 1] - 1[1(1-z) - 1] + 1[1 - (1-y)] = 0$. $(1-x)(1-z-y+yz-1) - (1-z-1) + (1-1+y) = 0$. $(1-x)(yz-y-z) + z + y = 0$. $yz - y - z - xyz + xy + xz + z + y = 0$. $yz + xy + xz - xyz = 0$. Dividing by $xyz$ (assuming $x, y, z 
eq 0$): $\frac{1}{x} + \frac{1}{z} + \frac{1}{y} = 1$. Thus,the locus is $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$.

Point	Plane
$(-6, 0, 0)$	$2x - 3y + 6z - 2 = 0$

If the points $A(2-x, 2, 2)$,$B(2, 2-y, 2)$,$C(2, 2, 2-z)$,and $D(1, 1, 1)$ are coplanar,then the locus of point $P(x, y, z)$ is

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