The equation of the plane which intersects the axis at \(\mathrm{A}, \mathrm{B}\), \(\mathrm{C}\) and the centroid of \(\Delta \mathrm{ABC}\) is \((\alpha, \beta, \gamma)\) is (A) \(x+y+z=3 \alpha \beta \gamma\) (B) \(x+y+z=3\) (C) \((\mathrm{x} / \alpha)+(\mathrm{y} / \beta)+(\mathrm{z} / \gamma)=3\) (D) \(\mathrm{x}+\mathrm{y}+\mathrm{z}=\alpha \beta \gamma\)

Notice that points A, B, and C are the points where the plane intersects the x, y, z-axes, respectively. Using this information, we can write the coordinates of the points: A: \((\alpha, 0, 0)\) B: \((0, \beta, 0)\) C: \((0, 0, \gamma)\)

The centroid of a triangle with vertices (a1, b1, c1), (a2, b2, c2), and (a3, b3, c3) is given by: \(\left( \frac{a1+a2+a3}{3}, \frac{b1+b2+b3}{3}, \frac{c1+c2+c3}{3} \right)\) Using this formula and taking into consideration the coordinates of A, B, and C, we can find the centroid of triangle ABC: \(\left( \frac{\alpha+0+0}{3}, \frac{0+\beta+0}{3}, \frac{0+0+\gamma}{3} \right) = \left( \frac{\alpha}{3}, \frac{\beta}{3}, \frac{\gamma}{3} \right)\)

Now that we have the coordinates of the centroid, let's analyze the given answer options to find the equation of the plane which contains points A, B, C, and the centroid. We can quickly eliminate options A and D, since they involve multiplying the coordinates, which is not the case for a plane equation involving the three axes. Now let's check option B (\(x+y+z = 3\)) by substituting the coordinates of the centroid \(\left(\frac{\alpha}{3}, \frac{\beta}{3}, \frac{\gamma}{3}\right)\): \(\frac{\alpha}{3}+\frac{\beta}{3}+\frac{\gamma}{3} = 3\) \(\frac{\alpha+\beta+\gamma}{3} = 3\) \(\alpha+\beta+\gamma = 9\) We find that option B doesn't provide the correct equation since it doesn't involve the coefficients \(\alpha, \beta,\) and \(\gamma\). Now let's check option C (\( \frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma} = 3\)) by substituting the coordinates of the centroid \(\left(\frac{\alpha}{3}, \frac{\beta}{3}, \frac{\gamma}{3}\right)\): \( \frac{\frac{\alpha}{3}}{\alpha}+\frac{\frac{\beta}{3}}{\beta}+\frac{\frac{\gamma}{3}}{\gamma} = 3\) \( \frac{1}{3}+\frac{1}{3}+\frac{1}{3} =1\) This equation satisfies the given condition, and therefore, the correct equation of the plane which intersects the axes at A, B, C, and the centroid of \(\Delta ABC \ ( \alpha, \beta , \gamma )\) is: \((x / \alpha) + (y / \beta) + (z / \gamma) = 3\) which corresponds to option C.