If the plane `(x)/(2)+(y)/(3)+(z)/(6)=1` cuts the coordinates axes at points A,B and C. Then, find the area of `triangle` ABC.

Options

A. `sqrt(18)` sq units
B. 30 sq units
C. `3sqrt(14)` sq units
D. `13sqrt(14)` sq units

Answer

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

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

1. **Intersection with the x-axis**: Set \(y = 0\) and \(z = 0\).
\[ \frac{x}{2} = 1 \implies x = 2 \]
Thus, the point A is \(A(2, 0, 0)\).

2. **Intersection with the y-axis**: Set \(x = 0\) and \(z = 0\).
\[ \frac{y}{3} = 1 \implies y = 3 \]
Thus, the point B is \(B(0, 3, 0)\).

3. **Intersection with the z-axis**: Set \(x = 0\) and \(y = 0\).
\[ \frac{z}{6} = 1 \implies z = 6 \]
Thus, the point C is \(C(0, 0, 6)\).

### Step 2: Write down the coordinates of points A, B, and C.
- \(A(2, 0, 0)\)
- \(B(0, 3, 0)\)
- \(C(0, 0, 6)\)

### Step 3: Find the vectors AB and AC.
- The vector \( \overrightarrow{AB} = B - A = (0 - 2, 3 - 0, 0 - 0) = (-2, 3, 0)\).
- The vector \( \overrightarrow{AC} = C - A = (0 - 2, 0 - 0, 6 - 0) = (-2, 0, 6)\).

### Step 4: Compute the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \).
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & 0 \\ -2 & 0 & 6 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 3 & 0 \\ 0 & 6 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & 0 \\ -2 & 6 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 3 \\ -2 & 0 \end{vmatrix} \]
\[ = \hat{i} (3 \cdot 6 - 0 \cdot 0) - \hat{j} (-2 \cdot 6 - 0 \cdot -2) + \hat{k} (-2 \cdot 0 - 3 \cdot -2) \]
\[ = 18\hat{i} + 12\hat{j} + 6\hat{k} \]

### Step 5: Find the magnitude of the cross product.
\[ |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(18)^2 + (12)^2 + (6)^2} = \sqrt{324 + 144 + 36} = \sqrt{504} \]

### Step 6: Calculate the area of triangle ABC.
The area of triangle ABC is given by:
\[ \text{Area} = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2} \sqrt{504} = \frac{\sqrt{36 \cdot 14}}{2} = \frac{6\sqrt{14}}{2} = 3\sqrt{14} \]

### Final Answer:
The area of triangle ABC is \(3\sqrt{14}\) square units.

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