The lines \[\frac{x-a+d}{\alpha -\delta }=\frac{y-a}{\alpha }=\frac{z-a-d}{\alpha +\delta }\] and \[\frac{x-b+c}{\beta -\gamma }=\frac{y-b}{\beta }=\frac{z-b-c}{\beta +\gamma }\] are coplanar and then equation to the plane in which they lie, is

The lines \[\frac{x-a+d}{\alpha -\delta }=\frac{y-a}{\alpha }=\frac{z-

1.	The lines \[\frac{x-a+d}{\alpha -\delta }=\frac{y-a}{\alpha }=\frac{z-a-d}{\alpha +\delta }\] and \[\frac{x-b+c}{\beta -\gamma }=\frac{y-b}{\beta }=\frac{z-b-c}{\beta +\gamma }\] are coplanar and then equation to the plane in which they lie, is
A.	\[x+y+z=0\]
B.	\[x-y+z=0\]
C.	\[x-2y+z=0\]
D.	\[x+y-2z=0\]
Answer» D. \[x+y-2z=0\]

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The lines \[\frac{x-a+d}{\alpha -\delta }=\frac{y-a}{\alpha }=\frac{z-a-d}{\alpha +\delta }\] and \[\frac{x-b+c}{\beta -\gamma }=\frac{y-b}{\beta }=\frac{z-b-c}{\beta +\gamma }\] are coplanar and then equation to the plane in which they lie, is

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