The intercepts of the plane `5x-3y+6z-60=0` on the coordinate axes are

Options

A. `10,20,-10`
B. `10,-20,12`
C. `12,-20,10`
D. `12,20,-10`

Explanation

To find the intercepts of the plane given by the equation \(5x - 3y + 6z - 60 = 0\) on the coordinate axes, we will convert the equation into intercept form and extract the intercepts accordingly.

### Step-by-Step Solution:

1. **Rearranging the Equation:**
Start with the given equation of the plane:
\[ 5x - 3y + 6z - 60 = 0 \]
Rearranging it gives:
\[ 5x - 3y + 6z = 60 \]

2. **Dividing by the Constant:**
To convert this into intercept form, we divide the entire equation by 60:
\[ \frac{5x}{60} - \frac{3y}{60} + \frac{6z}{60} = 1 \]
Simplifying each term results in:
\[ \frac{x}{12} - \frac{y}{20} + \frac{z}{10} = 1 \]

3. **Identifying the Intercepts:**
The intercept form of the plane is:
\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]
From our equation, we can identify:
- The x-intercept \(a = 12\)
- The y-intercept \(b = -20\)
- The z-intercept \(c = 10\)

4. **Conclusion:**
Therefore, the intercepts of the plane on the coordinate axes are:
- x-intercept: 12 units
- y-intercept: -20 units (indicating it is in the negative direction of the y-axis)
- z-intercept: 10 units

### Final Answer:
The intercepts of the plane on the coordinate axes are \( (12, -20, 10) \).