1. Key Concepts and Formulas

• Parametric Form of a Line: A line passing through a point $(x_0, y_0, z_0)$ with a direction vector $\vec{d} = (a, b, c)$ can be expressed in parametric form as $x = x_0 + \lambda a$, $y = y_0 + \lambda b$, $z = z_0 + \lambda c$, where $\lambda$ is a scalar parameter. This form simplifies finding coordinates of any point on the line and is crucial for determining intersection points.
• Intersection of Two Lines: To find the intersection point of two lines in 3D space, we equate their respective $x, y, z$ coordinates from their parametric forms. This generates a system of three linear equations with two parameters (one for each line). Solving this system yields the parameter values corresponding to the intersection point.
• Area of a Triangle using Vector Cross Product: Given three vertices of a triangle, $V_1, V_2, V_3$, its area $A$ can be calculated as half the magnitude of the cross product of two vectors forming two sides of the triangle originating from a common vertex. For example, $A = \frac{1}{2} |\vec{V_1V_2} \times \vec{V_1V_3}|$. This formula is derived from the geometric interpretation of the cross product, where its magnitude represents the area of the parallelogram formed by the two vectors.

2. Step-by-Step Solution

Step 1: Express the given lines in parametric form. We convert each symmetric equation of a line into its parametric form to easily represent any point on the line.

• Line $L_1$: $x+2=y-1=z$ Let $x+2=y-1=z = \lambda$. Then, $x = -2 + \lambda$, $y = 1 + \lambda$, $z = \lambda$. A point on $L_1$ is $P_1(-2, 1, 0)$ and its direction vector is $\vec{d_1} = (1, 1, 1)$.

• Line $L_2$: $\frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}$ Let $\frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1} = \mu$. Then, $x = 3 + 5\mu$, $y = -\mu$, $z = 1 + \mu$. A point on $L_2$ is $P_2(3, 0, 1)$ and its direction vector is $\vec{d_2} = (5, -1, 1)$.

• Line $L_3$: $\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}$ Let $\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1} = \nu$. Then, $x = -3\nu$, $y = 3 + 3\nu$, $z = 2 + \nu$. A point on $L_3$ is $P_3(0, 3, 2)$ and its direction vector is $\vec{d_3} = (-3, 3, 1)$.

Step 2: Find the intersection point $V_1$ of lines $L_1$ and $L_2$. We equate the coordinates of a general point on $L_1$ with a general point on $L_2$ to find the parameters $\lambda$ and $\mu$ that define their intersection. $-2 + \lambda = 3 + 5\mu \implies \lambda - 5\mu = 5 \quad (1)$ $1 + \lambda = -\mu \implies \lambda + \mu = -1 \quad (2)$ $\lambda = 1 + \mu \implies \lambda - \mu = 1 \quad (3)$ Subtracting equation (3) from equation (2): $(\lambda + \mu) - (\lambda - \mu) = -1 - 1 \implies 2\mu = -2 \implies \mu = -1$. Substitute $\mu = -1$ into equation (2): $\lambda + (-1) = -1 \implies \lambda = 0$. Check with equation (1): $0 - 5(-1) = 5$, which is consistent. Substitute $\lambda = 0$ into the parametric equations for $L_1$: $V_1 = (-2+0, 1+0, 0) = (-2, 1, 0)$.

Step 3: Find the intersection point $V_2$ of lines $L_2$ and $L_3$. We equate the coordinates of a general point on $L_2$ with a general point on $L_3$. $3 + 5\mu = -3\nu \implies 5\mu + 3\nu = -3 \quad (4)$ $-\mu = 3 + 3\nu \implies \mu + 3\nu = -3 \quad (5)$ $1 + \mu = 2 + \nu \implies \mu - \nu = 1 \quad (6)$ Subtracting equation (5) from equation (4): $(5\mu + 3\nu) - (\mu + 3\nu) = -3 - (-3) \implies 4\mu = 0 \implies \mu = 0$. Substitute $\mu = 0$ into equation (6): $0 - \nu = 1 \implies \nu = -1$. Check with equation (5): $0 + 3(-1) = -3$, which is consistent. Substitute $\mu = 0$ into the parametric equations for $L_2$: $V_2 = (3+5(0), -0, 1+0) = (3, 0, 1)$.

Step 4: Find the intersection point $V_3$ of lines $L_3$ and $L_1$. We equate the coordinates of a general point on $L_3$ with a general point on $L_1$. $-3\nu = -2 + \lambda \implies \lambda + 3\nu = 2 \quad (7)$ $3 + 3\nu = 1 + \lambda \implies \lambda - 3\nu = 2 \quad (8)$ $2 + \nu = \lambda \implies \lambda - \nu = 2 \quad (9)$ Subtracting equation (8) from equation (7): $(\lambda + 3\nu) - (\lambda - 3\nu) = 2 - 2 \implies 6\nu = 0 \implies \nu = 0$. Substitute $\nu = 0$ into equation (9): $\lambda - 0 = 2 \implies \lambda = 2$. Check with equation (8): $2 - 3(0) = 2$, which is consistent. Substitute $\nu = 0$ into the parametric equations for $L_3$: $V_3 = (-3(0), 3+3(0), 2+0) = (0, 3, 2)$.

The vertices of the triangle are $V_1 = (-2, 1, 0)$, $V_2 = (3, 0, 1)$, and $V_3 = (0, 3, 2)$.

Step 5: Calculate the vectors forming two sides of the triangle. We choose $V_1$ as the common starting point for the vectors. $\vec{V_1V_2} = V_2 - V_1 = (3 - (-2), 0 - 1, 1 - 0) = (5, -1, 1)$ $\vec{V_1V_3} = V_3 - V_1 = (0 - (-2), 3 - 1, 2 - 0) = (2, 2, 2)$

Step 6: Calculate the cross product of these two vectors. The cross product $\vec{N} = \vec{V_1V_2} \times \vec{V_1V_3}$ gives a vector whose magnitude is the area of the parallelogram formed by the two vectors. $\vec{N} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & -1 & 1 \\ 2 & 2 & 2 \end{vmatrix}$ $\vec{N} = \mathbf{i}((-1)(2) - (1)(2)) - \mathbf{j}((5)(2) - (1)(2)) + \mathbf{k}((5)(2) - (-1)(2))$ $\vec{N} = \mathbf{i}(-2 - 2) - \mathbf{j}(10 - 2) + \mathbf{k}(10 + 2)$ $\vec{N} = -4\mathbf{i} - 8\mathbf{j} + 12\mathbf{k} = (-4, -8, 12)$

Step 7: Find the magnitude of the cross product and calculate the area of the triangle. The magnitude of $\vec{N}$ is: $|\vec{N}| = \sqrt{(-4)^2 + (-8)^2 + (12)^2} = \sqrt{16 + 64 + 144} = \sqrt{224}$ The area $A$ of the triangle is half the magnitude of the cross product: $A = \frac{1}{2} |\vec{N}| = \frac{1}{2}\sqrt{224}$ We need to find $A^2$: $A^2 = \left(\frac{1}{2}\sqrt{224}\right)^2 = \frac{1}{4} \times 224 = 56$

3. Common Mistakes & Tips

• Arithmetic Errors: Calculating determinants and magnitudes involves many basic arithmetic operations. Double-check all additions, subtractions, and multiplications, especially with negative numbers.
• Incorrect Parametric Form: Ensure the symmetric form of the line is correctly converted to parametric form, especially when the denominator is 1 (e.g., $z = z_0 + \lambda \cdot 1$).
• System of Equations Errors: When solving for the parameters ($\lambda, \mu, \nu$), carefully solve the system of equations. Using elimination or substitution methods and verifying with the third equation is crucial.
• Vector Subtraction Order: Ensure consistency in vector subtraction (e.g., $\vec{V_1V_2} = V_2 - V_1$, not $V_1 - V_2$) to avoid sign errors in the cross product.

4. Summary

This problem involved finding the area of a triangle whose vertices are the pairwise intersection points of three lines in 3D space. We first converted the lines into their parametric forms to systematically find their intersection points. Once the three vertices of the triangle were identified, we formed two vectors representing two sides of the triangle originating from a common vertex. The area of the triangle was then calculated as half the magnitude of the cross product of these two vectors. Finally, we squared this area to obtain the required value. The calculated area squared is 56.

5. Final Answer

The final answer is $\boxed{56}$.