Question
The equation of the plane which makes with co-ordinate axes, a triangle with its centroid $$\left( {\alpha ,\,\beta ,\,\gamma } \right)$$ is :
A.
$$\alpha x + \beta y + \gamma z = 3$$
B.
$$\alpha x + \beta y + \gamma z = 1$$
C.
$$\frac{x}{\alpha } + \frac{y}{\beta } + \frac{z}{\gamma } = 3$$
D.
$$\frac{x}{\alpha } + \frac{y}{\beta } + \frac{z}{\gamma } = 1$$
Answer :
$$\frac{x}{\alpha } + \frac{y}{\beta } + \frac{z}{\gamma } = 3$$
Solution :
Let us take a triangle $$ABC$$ and their vertices $$A\left( {a,\,0,\,0} \right),\,B\left( {0,\,b,\,0} \right)$$ and $$C\left( {0,\,0,\,c} \right)$$
Therefore the equation of plane is
$$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1......\left( {\text{i}} \right)$$
Now, given centroid of $$\Delta ABC$$ is $$\left( {\alpha ,\,\beta ,\,\gamma } \right)$$
As we know, centroid of $$\Delta ABC$$ with vertices $$\left( {{x_1},\,{y_1},\,{z_1}} \right),\,\left( {{x_2},\,{y_2},\,{z_2}} \right)$$ and $$\left( {{x_3},\,{y_3},\,{z_3}} \right)$$ is given by
$$\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\,\frac{{{y_1} + {y_2} + {y_3}}}{3},\,\frac{{{z_1} + {z_2} + {z_3}}}{3}} \right)$$
$$\therefore $$ By using this formula, we have
$$\eqalign{
& \frac{{a + 0 + 0}}{3} = \alpha \, \Rightarrow a = 3\alpha \,; \cr
& \frac{{0 + b + 0}}{3} = \beta \, \Rightarrow b = 3\beta \,; \cr
& {\text{and }}\frac{{0 + 0 + c}}{3} = \gamma \, \Rightarrow c = 3\gamma \cr} $$
Now, put the values of $$a,\,b,\,c$$ in equation $$\left( {\text{i}} \right)$$,
which gives
$$\eqalign{
& \frac{x}{{3\alpha }} + \frac{y}{{3\beta }} + \frac{z}{{3\gamma }} = 1 \cr
& \therefore \,\frac{x}{\alpha } + \frac{y}{\beta } + \frac{z}{\gamma } = 3 \cr} $$