If the plane `x/2+y/3+z/6=1` cuts the axes of coordinates at points, `A ,B, ` and `C`, then find the area of the triangle `A B C`. a. `18` sq unit b. `36` sq unit c. `3sqrt(14)` sq unit d. `2sqrt(14)` sq unit

Options

A. 18 sq unit
B. 36 sq unit
C. `3sqrt14" sq unit"`
D. `2sqrt14" sq unit"`

Explanation

To find the area of the triangle formed by the points where the plane \( \frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1 \) intersects the coordinate axes, we follow these steps:

### Step 1: Find the points of intersection with the axes

The plane intersects the axes at points \( A \), \( B \), and \( C \).

1. **Point A (x-axis)**: Set \( y = 0 \) and \( z = 0 \):
\[ \frac{x}{2} = 1 \implies x = 2 \implies A(2, 0, 0) \]

2. **Point B (y-axis)**: Set \( x = 0 \) and \( z = 0 \):
\[ \frac{y}{3} = 1 \implies y = 3 \implies B(0, 3, 0) \]

3. **Point C (z-axis)**: Set \( x = 0 \) and \( y = 0 \):
\[ \frac{z}{6} = 1 \implies z = 6 \implies C(0, 0, 6) \]

### Step 2: Determine the vectors AB and AC

Now, we find the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \).

1. **Vector \( \overrightarrow{AB} \)**:
\[ \overrightarrow{AB} = B - A = (0 - 2, 3 - 0, 0 - 0) = (-2, 3, 0) \]

2. **Vector \( \overrightarrow{AC} \)**:
\[ \overrightarrow{AC} = C - A = (0 - 2, 0 - 0, 6 - 0) = (-2, 0, 6) \]

### Step 3: Calculate the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \)

To find the area of triangle \( ABC \), we need the magnitude of the cross product of vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \).

\[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & 0 \\ -2 & 0 & 6 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 3 & 0 \\ 0 & 6 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & 0 \\ -2 & 6 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 3 \\ -2 & 0 \end{vmatrix} \]

Calculating each of these determinants:

1. \( \hat{i}(3 \cdot 6 - 0 \cdot 0) = 18\hat{i} \)

2. \( -\hat{j}((-2) \cdot 6 - 0 \cdot (-2)) = -(-12)\hat{j} = 12\hat{j} \)

3. \( \hat{k}((-2) \cdot 0 - 3 \cdot (-2)) = 6\hat{k} \)

Combining these, we have:

\[ \overrightarrow{AB} \times \overrightarrow{AC} = 18\hat{i} + 12\hat{j} + 6\hat{k} \]

### Step 4: Calculate the magnitude of the cross product

The magnitude is given by:

\[ |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{18^2 + 12^2 + 6^2} = \sqrt{324 + 144 + 36} = \sqrt{504} \]

### Step 5: Calculate the area of triangle ABC

The area \( \Delta \) of triangle \( ABC \) is given by:

\[ \Delta = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2} \sqrt{504} = \frac{1}{2} \sqrt{36 \times 14} = \frac{6\sqrt{14}}{2} = 3\sqrt{14} \]

Thus, the area of triangle \( ABC \) is \( 3\sqrt{14} \) square units.

### Final Answer

The area of triangle \( ABC \) is \( \boxed{3\sqrt{14}} \) square units.