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**Hint:**A triangle ABC is placed so that the mid-points of the sides are on the x,y,z axes. Lengths of the intercepts made by the plane containing the triangle on these axes are respectively\[\alpha ,\beta ,\gamma \]. Coordinates of the centroid of the triangle ABC are

\[( - \alpha /3,\beta /3,\gamma /3)\]

\[(\alpha /3, - \beta /3,\gamma /3)\]

\[(\alpha /3,\beta /3, - \gamma /3)\]

\[(\alpha /3,\beta /3,\gamma /3)\]

**Complete step by step solution:**

1) \[\,ABC\] is placed so that the mid-points of the sides are on the x,y,z axes

i.e.:D,E,F are the mid-points of CA , BC , BA respectively and D,E,F lie on the Z,Y,X axes.

2) Lengths of the intercepts made by the plane containing the triangle on these axes are \[\alpha ,\beta ,\gamma \]respectively which means the coordinates of D,E,F are \[F(\alpha ,0,0),E(0,\beta ,0)and\,D(0,0,\gamma )\].

Here, \[A = ({x_1},{y_1},{z_1})\]

\[B = ({x_2},{y_2},{z_2})\]

\[C = ({x_3},{y_3},{z_3})\]

And \[F = (\alpha ,0,0)\]

\[E = (0,\beta ,0)\]

\[D = (0,0,\gamma )\]

Step 1: To make a relation between the coordinates of A,B,C with \[\alpha ,\beta ,\gamma \]

i)= > \[F = (\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2},\dfrac{{{z_1} + {z_2}}}{2})\] [ F is the mid point of AB]

= > \[F = (\alpha ,0,0)\]

i.e.: \[\dfrac{{{x_1} + {x_2}}}{2} = \alpha ,\,\,\dfrac{{{y_1} + {y_2}}}{2} = 0,\,\,\dfrac{{{z_1} + {z_2}}}{2} = 0\]

ii) = > \[E = (\dfrac{{{x_2} + {x_3}}}{2},\dfrac{{{y_2} + {y_3}}}{2},\dfrac{{{z_2} + {z_3}}}{2})\] [ E is the mid point of BC]

= > \[F = (0,\beta ,0)\]

i.e.: \[\dfrac{{{x_2} + {x_3}}}{2} = 0,\,\,\dfrac{{{y_2} + {y_3}}}{2} = \beta ,\,\,\dfrac{{{z_2} + {z_3}}}{2} = 0\]

=> \[D = (\dfrac{{{x_3} + {x_1}}}{2},\dfrac{{{y_3} + {y_1}}}{2},\dfrac{{{z_3} + {z_1}}}{2})\]

= > \[D = (0,0,\gamma )\]

i.e.: \[\dfrac{{{x_3} + {x_1}}}{2} = 0,\,\,\dfrac{{{y_3} + {y_1}}}{2} = 0,\,\,\dfrac{{{z_3} + {z_1}}}{2} = \gamma \]

From (i) , We get,

\[{x_1} + {x_2} = 2\alpha \]

\[{y_1} + {y_2} = 0\]

\[{z_1} + {z_2} = 0\]

From (ii), we get,

\[{x_2} + {x_3} = 0\]

\[{y_2} + {y_3} = 2\beta \]

\[{z_2} + {z_3} = 0\]

From (iii), we get,

\[{x_3} + {x_1} = 0\]

\[{y_3} + {y_1} = 0\]

\[{z_3} + {z_1} = 2\gamma \]

From all the above equations, we can conclude that,

\[{x_1} + {x_2} + {x_3} = \alpha \]

\[{y_1} + {y_2} + {y_3} = \beta \]

\[{z_1} + {z_2} + {z_3} = \gamma \]

Hence, we get the coordinates of the points A,B,C in terms of \[\alpha ,\beta ,\gamma \]

Step 2: We will now find out the coordinate of centroid of the triangle coordinate of the centroid of the triangle is given by

\[(\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3},\dfrac{{{z_1} + {z_2} + {z_3}}}{3})\]

= \[(\dfrac{\alpha }{3},\dfrac{\beta }{3},\dfrac{\gamma }{3})\]

Hence, the coordinate of the centroid of the triangle ABC is \[(\dfrac{\alpha }{3},\dfrac{\beta }{3},\dfrac{\gamma }{3})\].

So, \[D(\dfrac{\alpha }{3},\dfrac{\beta }{3},\dfrac{\gamma }{3})\] is the correct answer.

**Note:**Diagram should be drawn properly. As because, a correct diagram will only lead you to a correct answer. Also, equations should be written correctly.

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