1. Key Concepts and Formulas

• Area of a Triangle: The area of any triangle can be calculated using the formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
• Perpendicular Distance (Altitude): In a triangle, the height corresponding to a base is the perpendicular distance from the opposite vertex to the line containing that base.
• Perpendicularity Condition in 3D: If a vector $\vec{v_1}$ is perpendicular to a vector $\vec{v_2}$, their dot product is zero ($\vec{v_1} \cdot \vec{v_2} = 0$). This condition is used to find the foot of the perpendicular from a point to a line.
• Distance Formula in 3D: The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$.

2. Step-by-Step Solution

Step 1: Parameterize the Line Containing Vertices B and C The line containing vertices $B$ and $C$ is given by the symmetric form: ${{x + 2} \over 3} = {{y - 1} \over 0} = {z \over 4}$ To represent any general point on this line, we set each part equal to a scalar parameter, say $\lambda$. From the equation, we get: $x + 2 = 3\lambda \quad \Rightarrow \quad x = 3\lambda - 2$ $y - 1 = 0 \cdot \lambda \quad \Rightarrow \quad y = 1$ $z = 4\lambda \quad \Rightarrow \quad z = 4\lambda$ So, any point $D$ on the line $BC$ can be represented by the coordinates: $D(3\lambda - 2, 1, 4\lambda)$ The direction ratios (DRs) of the line $BC$ are given by the denominators of the symmetric form, which are $\langle 3, 0, 4 \rangle$. This vector $\vec{d} = 3\hat{i} + 0\hat{j} + 4\hat{k}$ represents the direction of the line.

Step 2: Define the Vector AD and its Direction Ratios We are given the coordinates of point $A(1, -1, 2)$. The coordinates of a general point $D$ on the line $BC$ are $(3\lambda - 2, 1, 4\lambda)$. The vector $\vec{AD}$ is found by subtracting the coordinates of $A$ from $D$: $\vec{AD} = ( (3\lambda - 2) - 1, \quad 1 - (-1), \quad 4\lambda - 2 )$ $\vec{AD} = ( 3\lambda - 3, \quad 2, \quad 4\lambda - 2 )$ The direction ratios of vector $\vec{AD}$ are $\langle 3\lambda - 3, 2, 4\lambda - 2 \rangle$. This vector represents the altitude from $A$ to the line $BC$ when $D$ is the foot of the perpendicular.

Step 3: Apply the Perpendicularity Condition Since $D$ is the foot of the perpendicular from $A$ to the line $BC$, the vector $\vec{AD}$ must be perpendicular to the line $BC$. For two vectors to be perpendicular, their dot product must be zero. Direction ratios of $\vec{AD}$: $\langle 3\lambda - 3, 2, 4\lambda - 2 \rangle$ Direction ratios of line $BC$: $\langle 3, 0, 4 \rangle$ Applying the dot product condition: $(3\lambda - 3)(3) + (2)(0) + (4\lambda - 2)(4) = 0$ This equation will allow us to find the specific value of $\lambda$ that corresponds to the foot of the perpendicular, $D$.

Step 4: Solve for the Parameter $\lambda$ and Find Coordinates of D Now, we solve the equation obtained in Step 3 for $\lambda$: $9\lambda - 9 + 0 + 16\lambda - 8 = 0$ $25\lambda - 17 = 0$ $25\lambda = 17$ $\lambda = \frac{17}{25}$ Substitute this value of $\lambda$ back into the coordinates of $D(3\lambda - 2, 1, 4\lambda)$ to find the exact coordinates of the foot of the perpendicular: $D_x = 3\left(\frac{17}{25}\right) - 2 = \frac{51}{25} - \frac{50}{25} = \frac{1}{25}$ $D_y = 1$ $D_z = 4\left(\frac{17}{25}\right) = \frac{68}{25}$ So, the coordinates of $D$ are $\left( \frac{1}{25}, 1, \frac{68}{25} \right)$.

Step 5: Calculate the Length of the Altitude AD Now we calculate the distance between $A(1, -1, 2)$ and $D\left( \frac{1}{25}, 1, \frac{68}{25} \right)$ using the 3D distance formula: $AD = \sqrt{(x_D - x_A)^2 + (y_D - y_A)^2 + (z_D - z_A)^2}$ $AD = \sqrt{\left(\frac{1}{25} - 1\right)^2 + (1 - (-1))^2 + \left(\frac{68}{25} - 2\right)^2}$ $AD = \sqrt{\left(\frac{1 - 25}{25}\right)^2 + (1 + 1)^2 + \left(\frac{68 - 50}{25}\right)^2}$ $AD = \sqrt{\left(\frac{-24}{25}\right)^2 + (2)^2 + \left(\frac{18}{25}\right)^2}$ $AD = \sqrt{\frac{576}{625} + 4 + \frac{324}{625}}$ To combine these terms, find a common denominator (625): $AD = \sqrt{\frac{576}{625} + \frac{4 \times 625}{625} + \frac{324}{625}}$ $AD = \sqrt{\frac{576 + 2500 + 324}{625}}$ $AD = \sqrt{\frac{3600}{625}}$ Now, simplify the square root: $AD = \frac{\sqrt{3600}}{\sqrt{625}} = \frac{60}{25}$ $AD = \frac{12}{5}$ So, the length of the altitude $AD$ is $\frac{12}{5}$ units.

Step 6: Calculate the Area of Triangle ABC Finally, we use the area formula with the given base $BC = 5$ units and the calculated height $AD = \frac{12}{5}$ units: $\text{Area of } \Delta ABC = \frac{1}{2} \times BC \times AD$ $\text{Area of } \Delta ABC = \frac{1}{2} \times 5 \times \frac{12}{5}$ $\text{Area of } \Delta ABC = \frac{1}{2} \times 12$ $\text{Area of } \Delta ABC = 6$ The area of the triangle is 6 square units.

3. Common Mistakes & Tips

• Arithmetic Errors: Be extremely careful with fraction arithmetic, especially when squaring and finding common denominators. Even small calculation mistakes can lead to an incorrect final answer.
• Perpendicularity Condition: Ensure the dot product is correctly applied using the direction ratios of the altitude vector and the line's direction vector. A common mistake is using the coordinates of a point on the line instead of its direction ratios.
• Distance Formula: Double-check the signs and squaring operations when calculating distances in 3D.

4. Summary

To find the area of the triangle, we first identified the base length $BC = 5$ units. The main challenge was to find the height, which is the perpendicular distance from vertex $A$ to the line containing $BC$. We achieved this by parameterizing the line, forming a vector $\vec{AD}$ (where $D$ is a point on the line), and then using the condition that $\vec{AD}$ is perpendicular to the line. This allowed us to find the coordinates of $D$, the foot of the perpendicular, and subsequently calculate the length of $AD$. Finally, the standard area formula $\frac{1}{2} \times \text{base} \times \text{height}$ yielded the area of the triangle.

The final answer is $\boxed{6}$, which corresponds to option (A).