Question

If a plane meets the axes $X$, $Y$, $Z$ in $A$, $B$, $C$ respectively such that centroid of $\Delta ABC$ is $(1, 2, 3)$, then the equation of the plane is

• A. $x + 2y + 3z = 1$
• B. $x + \frac{y}{2} + \frac{z}{3} = 3$
• C. $\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1$
• D. $\frac{x}{4} + \frac{y}{8} + \frac{z}{12} = 1$

Hint:

For any plane whose axes-intercept triangle has a centroid $(x_0, y_0, z_0)$, the values of the intercepts are always exactly three times the centroid's coordinates ($a = 3x_0, b = 3y_0, c = 3z_0$). This allows you to write down the final equation $\frac{x}{3x_0} + \frac{y}{3y_0} + \frac{z}{3z_0} = 1$ in a single step!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A plane intersects the three coordinate axes at points $A$, $B$, and $C$, forming a triangle. Given the specific coordinates of the centroid of $\Delta ABC$, we need to find the Cartesian equation of the plane.

Step 2: Key Formula or Approach:
1. Let the intercepts of the plane on the $X$, $Y$, and $Z$ axes be $a$, $b$, and $c$ respectively. The vertices are $A(a, 0, 0)$, $B(0, b, 0)$, and $C(0, 0, c)$. 2. The coordinates of the centroid $G$ of a triangle with these vertices are given by: $$G = \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$$ 3. The intercept form equation of a plane is: $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$

Step 3: Detailed Explanation:
Equating the general centroid formula to the given centroid point $(1, 2, 3)$: $$\frac{a}{3} = 1 \implies a = 3$$ $$\frac{b}{3} = 2 \implies b = 6$$ $$\frac{c}{3} = 3 \implies c = 9$$ Now substitute these values for the intercepts $a$, $b$, and $c$ directly into the intercept form equation of the plane: $$\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1$$

Step 4: Final Answer:
The Cartesian equation of the plane is $\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1$, which matches option (C).