The plane `x / 2 + y / 3 + z / 4 = 1` cuts the co-ordinate axes in `A, B, C`: then the area of the `DeltaABC` is

Options

A. `sqrt(31)`
B. `sqrt(61)`
C. `sqrt(41)`
D. `sqrt(51)`

Answer

To find the area of triangle ABC formed by the intercepts of the plane \( \frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1 \) with the coordinate axes, we can follow these steps:

### Step 1: Determine the intercepts on the axes
The given equation of the plane is in the intercept form:
\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]
where \( a, b, c \) are the intercepts on the x, y, and z axes respectively.

From the equation \( \frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1 \), we can identify:
- \( a = 2 \) (x-intercept)
- \( b = 3 \) (y-intercept)
- \( c = 4 \) (z-intercept)

Thus, the intercepts are:
- Point A (2, 0, 0)
- Point B (0, 3, 0)
- Point C (0, 0, 4)

### Step 2: Use the formula for the area of triangle formed by intercepts
The area \( A \) of triangle ABC formed by the intercepts can be calculated using the formula:
\[ A = \frac{1}{2} \sqrt{a^2 + b^2 + c^2} \]
where \( a, b, c \) are the intercepts on the x, y, and z axes.

### Step 3: Substitute the values of \( a, b, c \)
Substituting the values \( a = 2 \), \( b = 3 \), and \( c = 4 \) into the formula:
\[ A = \frac{1}{2} \sqrt{2^2 + 3^2 + 4^2} \]

Calculating the squares:
\[ A = \frac{1}{2} \sqrt{4 + 9 + 16} \]
\[ A = \frac{1}{2} \sqrt{29} \]

### Step 4: Final area calculation
Thus, the area of triangle ABC is:
\[ A = \frac{1}{2} \sqrt{29} \text{ square units} \]

### Summary
The area of triangle ABC formed by the intercepts of the plane with the coordinate axes is:
\[ \frac{1}{2} \sqrt{29} \text{ square units} \]