Key Concepts and Formulas

• Parametric Equation of a Line: A line passing through a point $(x_0, y_0, z_0)$ with a direction vector $\langle a, b, c \rangle$ can be represented parametrically as $x = x_0 + a\lambda$, $y = y_0 + b\lambda$, $z = z_0 + c\lambda$, where $\lambda$ is a scalar parameter. This form is crucial for representing any general point on a line.
• Direction Vector of a Line: The direction vector of a line passing through two points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ is given by $\vec{P_1P_2} = \langle x_2-x_1, y_2-y_1, z_2-z_1 \rangle$. Also, for a line in symmetric form $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$, the direction vector is $\langle a, b, c \rangle$.
• Parallel Lines: Two lines are parallel if and only if their direction vectors are proportional. That is, if $\vec{d_1}$ and $\vec{d_2}$ are the direction vectors of two parallel lines, then $\vec{d_1} = k \cdot \vec{d_2}$ for some non-zero scalar $k$.
• Equation of a Line (Point-Direction Form): Given a point $(x_0, y_0, z_0)$ on the line and its direction vector $\langle a, b, c \rangle$, the symmetric equation of the line is $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$.

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Step-by-Step Solution

Step 1: Convert the given lines into their parametric forms. To find the line L, we first need to express the two lines it intersects ($L_1$ and $L_2$) in a way that allows us to represent any point on them using a parameter. This is best done using parametric equations.

• Line $L_1$: $x-2 = -y = z-1$ First, write it in standard symmetric form: $L_1: \frac{x-2}{1} = \frac{y-0}{-1} = \frac{z-1}{1}$ Now, set each ratio equal to a parameter, say $\lambda$, to get the parametric form for any point $P_1$ on $L_1$: $P_1(\lambda) = (\lambda+2, -\lambda, \lambda+1)$ Reasoning: The parametric form allows us to express the coordinates of any point on $L_1$ using a single variable $\lambda$. This is essential for defining the intersection point of $L$ with $L_1$.

• Line $L_2$: $2(x+1) = 2(y-1) = z+1$ To convert to standard symmetric form, we divide the first two parts by 2: $L_2: \frac{x+1}{1/2} = \frac{y-1}{1/2} = \frac{z+1}{1}$ Set each ratio equal to a parameter, say $\mu$, to get the parametric form for any point $P_2$ on $L_2$: $P_2(\mu) = \left(\frac{\mu}{2}-1, \frac{\mu}{2}+1, \mu-1\right)$ Reasoning: Similar to $L_1$, this parametric representation for $L_2$ will allow us to define the intersection point of $L$ with $L_2$.

Step 2: Determine the direction vector of line L. Line L intersects $L_1$ at a point $P_1(\lambda)$ and $L_2$ at a point $P_2(\mu)$. Therefore, the vector connecting these two points, $\vec{P_1P_2}$, must be a direction vector for line L. We calculate $\vec{P_1P_2} = P_2 - P_1$: $\vec{P_1P_2} = \left\langle \left(\frac{\mu}{2}-1\right) - (\lambda+2), \left(\frac{\mu}{2}+1\right) - (-\lambda), (\mu-1) - (\lambda+1) \right\rangle$ $\vec{P_1P_2} = \left\langle \frac{\mu}{2} - \lambda - 3, \frac{\mu}{2} + \lambda + 1, \mu - \lambda - 2 \right\rangle$ Reasoning: If a line passes through two specific points, the vector connecting these two points lies along the line and thus serves as its direction vector.

Step 3: Apply the parallelism condition. Line L is given to be parallel to the line $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$. The direction vector of this parallel line (let's call it $L_P$) is $\vec{d_P} = \langle 3, 1, 2 \rangle$. Since line L is parallel to $L_P$, their direction vectors must be proportional. This means $\vec{P_1P_2}$ must be a scalar multiple of $\vec{d_P}$. Let $k$ be the constant of proportionality: $\vec{P_1P_2} = k \cdot \vec{d_P}$ Equating the components gives us a system of proportions: $\frac{\frac{\mu}{2} - \lambda - 3}{3} = \frac{\frac{\mu}{2} + \lambda + 1}{1} = \frac{\mu - \lambda - 2}{2}$ Reasoning: This step translates the geometric condition of parallelism into algebraic equations involving our parameters $\lambda$ and $\mu$, which we can then solve.

Step 4: Solve the system of equations for $\lambda$ and $\mu$. From the proportionality in Step 3, we can form two independent linear equations:

• Equation (1): Equating the first two ratios: $\frac{\mu}{2} - \lambda - 3 = 3 \left( \frac{\mu}{2} + \lambda + 1 \right)$ $\frac{\mu}{2} - \lambda - 3 = \frac{3\mu}{2} + 3\lambda + 3$ Multiply by 2 to clear fractions: $\mu - 2\lambda - 6 = 3\mu + 6\lambda + 6$ Rearrange terms: $-12 = 2\mu + 8\lambda$ $\mu + 4\lambda = -6 \quad \text{(Eq. A)}$

• Equation (2): Equating the second and third ratios: $\frac{\mu}{2} + \lambda + 1 = \frac{\mu - \lambda - 2}{2}$ Multiply by 2 to clear fractions: $\mu + 2\lambda + 2 = \mu - \lambda - 2$ Rearrange terms: $2\lambda + 2 = -\lambda - 2$ $3\lambda = -4$ $\lambda = -\frac{4}{3}$

Now, substitute the value of $\lambda$ into Eq. A: $\mu + 4\left(-\frac{4}{3}\right) = -6$ $\mu - \frac{16}{3} = -6$ $\mu = -6 + \frac{16}{3}$ $\mu = -\frac{18}{3} + \frac{16}{3}$ $\mu = -\frac{2}{3}$ Reasoning: Solving these equations gives us the unique values of $\lambda$ and $\mu$ that define the specific intersection points $P_1$ and $P_2$ on $L_1$ and $L_2$ respectively, such that the line connecting them is parallel to $L_P$.

Step 5: Determine the equation of line L. We have found $\lambda = -\frac{4}{3}$ and $\mu = -\frac{2}{3}$. We can use either $P_1$ or $P_2$ as a point on line L. Let's use $P_1$. Substitute $\lambda = -\frac{4}{3}$ into the parametric form of $P_1(\lambda)$: $P_1 = \left(-\frac{4}{3}+2, -(-\frac{4}{3}), -\frac{4}{3}+1\right)$ $P_1 = \left(\frac{-4+6}{3}, \frac{4}{3}, \frac{-4+3}{3}\right)$ $P_1 = \left(\frac{2}{3}, \frac{4}{3}, -\frac{1}{3}\right)$ The direction vector of line L is $\vec{d_L} = \vec{d_P} = \langle 3, 1, 2 \rangle$. Using point $P_1$ and direction vector $\vec{d_L}$, the symmetric equation of line L is: $L: \frac{x - \frac{2}{3}}{3} = \frac{y - \frac{4}{3}}{1} = \frac{z - (-\frac{1}{3})}{2}$ $L: \frac{x - \frac{2}{3}}{3} = \frac{y - \frac{4}{3}}{1} = \frac{z + \frac{1}{3}}{2}$ Reasoning: With a specific point on the line and its direction vector, we can uniquely write the equation of the line.

Step 6: Check the given options. To find which point lies on line L, we substitute the coordinates of each option into the equation of L and check if all three ratios are equal.

Let's check option (A) $\left(-\frac{1}{3}, 1, -1\right)$: Substitute $x = -\frac{1}{3}, y = 1, z = -1$ into the equation of L: $\frac{-\frac{1}{3} - \frac{2}{3}}{3} = \frac{1 - \frac{4}{3}}{1} = \frac{-1 + \frac{1}{3}}{2}$ $\frac{-\frac{3}{3}}{3} = \frac{\frac{3-4}{3}}{1} = \frac{\frac{-3+1}{3}}{2}$ $\frac{-1}{3} = -\frac{1}{3} = \frac{-\frac{2}{3}}{2}$ $-\frac{1}{3} = -\frac{1}{3} = -\frac{1}{3}$ All three ratios are equal to $-\frac{1}{3}$. Therefore, point (A) lies on line L. Reasoning: A point lies on a line if and only if its coordinates satisfy the line's equation.

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Common Mistakes & Tips

• Converting to Standard Form: Be extremely careful when converting lines like $x-2=-y=z-1$ or $2(x+1)=2(y-1)=z+1$ to standard symmetric form. Pay attention to signs and coefficients (e.g., $-y$ becomes $\frac{y-0}{-1}$, and $2(x+1)$ becomes $\frac{x+1}{1/2}$ or, by multiplying through by 2, $\frac{x+1}{1}$).
• Algebraic Errors: The most common errors occur during the solution of the system of linear equations for $\lambda$ and $\mu$. Fractions often complicate calculations, so double-check each step.
• Choosing the Correct Direction Vector: Remember that the direction vector of line L is given by the line it's parallel to ($\vec{d_P}$), not necessarily the vector $\vec{P_1P_2}$ itself, although $\vec{P_1P_2}$ is proportional to $\vec{d_P}$. Using $\vec{d_P}$ ensures the simplest form for the direction vector.

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Summary

This problem combined several fundamental concepts of 3D geometry to find the equation of a specific line. We began by converting the intersecting lines into parametric forms to represent general points on them. Then, we formed a vector connecting these general points, which served as a direction vector for the desired line. By applying the parallelism condition with the third given line, we set up and solved a system of equations to find the precise values of the parameters. These parameters allowed us to determine a specific point on the line L. Finally, using this point and the known direction vector, we wrote the equation of line L and verified which of the given options lay on it. This systematic approach ensures all conditions are met, leading to the correct answer.

The final answer is $\boxed{\left(-\frac{1}{3}, 1,-1\right)}$, which corresponds to option (A).