A) \[\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3\]
B) \[\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1\]
C) \[\frac{3x}{\alpha }+\frac{3y}{\beta }+\frac{3z}{\gamma }=1\]
D) \[\alpha x+\beta y+\gamma z=1\]
Correct Answer: A
Solution :
Let the co-ordinates of the points where the plane cuts the axes are (a, 0, 0), (0, b, 0), (0, 0, c). Since centroid is \[(\alpha ,\,\,\beta ,\,\,\gamma ),\] therefore \[a=3\alpha ,\,\] \[b=3\beta ,\,\,c=3\gamma .\] Equation of the plane will be \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\] \[\Rightarrow \,\,\frac{x}{3\alpha }+\frac{y}{3\beta }+\frac{z}{3\gamma }=1\,\,\Rightarrow \,\,\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3.\]
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