Find the equation of the plane which cuts intercepts 2,3,-4 on the axes.

Options

A. `6x+4y-3z=12`
B. `6x-4y+3z=12`
C. `2x-3y+4z=12`
D. None of these

Answer

To find the equation of the plane that intercepts the axes at points \(2\), \(3\), and \(-4\), we can use the intercept form of the equation of a plane.

### Step-by-Step Solution:

1. **Understand the Intercept Form of the Plane:**
The intercept form of the equation of a plane is given by:
\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]
where \(a\), \(b\), and \(c\) are the x-intercept, y-intercept, and z-intercept, respectively.

2. **Identify the Intercepts:**
From the problem, we have:
- \(a = 2\) (x-intercept)
- \(b = 3\) (y-intercept)
- \(c = -4\) (z-intercept)

3. **Substitute the Intercepts into the Equation:**
Substitute \(a\), \(b\), and \(c\) into the intercept form:
\[ \frac{x}{2} + \frac{y}{3} + \frac{z}{-4} = 1 \]

4. **Simplify the Equation:**
To eliminate the fractions, we can multiply through by the least common multiple (LCM) of the denominators, which is 12:
\[ 12 \left(\frac{x}{2}\right) + 12 \left(\frac{y}{3}\right) + 12 \left(\frac{z}{-4}\right) = 12 \]
This simplifies to:
\[ 6x + 4y - 3z = 12 \]

5. **Final Equation of the Plane:**
The equation of the plane is:
\[ 6x + 4y - 3z - 12 = 0 \]