Key Concepts and Formulas

1. Parametric Equation of a Line: A line given by $\frac{x-x_0}{l}=\frac{y-y_0}{m}=\frac{z-z_0}{n}$ can be represented parametrically. Any point on this line can be expressed as $(x_0 + l\lambda, y_0 + m\lambda, z_0 + n\lambda)$, where $\lambda$ is a scalar parameter. The vector $\vec{d} = (l, m, n)$ is the direction vector of the line.
2. Foot of the Perpendicular: If $R$ is the foot of the perpendicular from a point $P$ to a line $L$, then the vector $\vec{PR}$ is perpendicular to the direction vector of the line $L$. This means their dot product is zero: $\vec{PR} \cdot \vec{d} = 0$.
3. Distance Formula in 3D: The distance between two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ is $AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$. This can also be seen as the magnitude of the vector $\vec{AB}$.
4. Image of a Point in a Line: If $Q$ is the image of a point $P$ in a line $L$, and $R$ is the foot of the perpendicular from $P$ to $L$, then $R$ is the midpoint of the line segment $PQ$. Thus, $R = \left( \frac{x_P+x_Q}{2}, \frac{y_P+y_Q}{2}, \frac{z_P+z_Q}{2} \right)$.

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Step-by-Step Solution

Step 1: Represent a general point on the line and form the vector $\vec{PR}$. Let the given point be $P(a, 4, 2)$. The given line $L$ is $\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$. We want to find the foot of the perpendicular from $P$ to $L$. Let this foot be $R$. Since $R$ lies on the line $L$, we can express its coordinates parametrically by setting each part of the line equation equal to $\lambda$: $\frac{x+1}{2} = \lambda \Rightarrow x = 2\lambda - 1$ $\frac{y-3}{3} = \lambda \Rightarrow y = 3\lambda + 3$ $\frac{z-1}{-1} = \lambda \Rightarrow z = -\lambda + 1$ So, any point on the line $L$, including $R$, can be represented as $R(2\lambda - 1, 3\lambda + 3, -\lambda + 1)$.

Now, we form the vector $\vec{PR}$ by subtracting the coordinates of $P$ from $R$: $\vec{PR} = ( (2\lambda - 1) - a, (3\lambda + 3) - 4, (-\lambda + 1) - 2 )$ $\vec{PR} = (2\lambda - 1 - a, 3\lambda - 1, -\lambda - 1)$ Reasoning: Representing a general point $R$ on the line parametrically is the first step to finding the specific point that is the foot of the perpendicular. The vector $\vec{PR}$ connects the given point $P$ to any point $R$ on the line.

Step 2: Use the perpendicularity condition to establish a relationship between $a$ and $\lambda$. The vector $\vec{PR}$ must be perpendicular to the direction vector of the line $L$ for $R$ to be the foot of the perpendicular. The direction vector of line $L$ is $\vec{d} = (2, 3, -1)$ (from the denominators of the symmetric equation).

For $\vec{PR}$ to be perpendicular to $\vec{d}$, their dot product must be zero: $\vec{PR} \cdot \vec{d} = 0$ $(2\lambda - 1 - a)(2) + (3\lambda - 1)(3) + (-\lambda - 1)(-1) = 0$ Expanding and simplifying: $4\lambda - 2 - 2a + 9\lambda - 3 + \lambda + 1 = 0$ $(4\lambda + 9\lambda + \lambda) + (-2 - 3 + 1) - 2a = 0$ $14\lambda - 4 - 2a = 0$ Dividing by 2: $7\lambda - 2 - a = 0$ This gives us a crucial relationship: $a = 7\lambda - 2 \quad \text{(Equation 1)}$ Reasoning: This condition allows us to find the unique value of $\lambda$ (and thus $R$) that makes $PR$ perpendicular to the line.

Step 3: Utilize the given perpendicular distance to form a quadratic equation in $\lambda$. We are given that the length of the perpendicular $PR$ is $2\sqrt{6}$ units. Therefore, $PR^2 = (2\sqrt{6})^2 = 4 \times 6 = 24$. Using the distance formula (magnitude of $\vec{PR}$): $PR^2 = (2\lambda - 1 - a)^2 + (3\lambda - 1)^2 + (-\lambda - 1)^2 = 24$ Substitute $a = 7\lambda - 2$ from Equation 1 into the first component of $\vec{PR}$: $(2\lambda - 1 - a) = 2\lambda - 1 - (7\lambda - 2) = 2\lambda - 1 - 7\lambda + 2 = -5\lambda + 1$ Now substitute this back into the $PR^2$ equation: $(-5\lambda + 1)^2 + (3\lambda - 1)^2 + (-\lambda - 1)^2 = 24$ Note that $(-\lambda - 1)^2 = (\lambda + 1)^2$. Expanding each squared term: $(25\lambda^2 - 10\lambda + 1) + (9\lambda^2 - 6\lambda + 1) + (\lambda^2 + 2\lambda + 1) = 24$ Combine like terms: $(25+9+1)\lambda^2 + (-10-6+2)\lambda + (1+1+1) = 24$ $35\lambda^2 - 14\lambda + 3 = 24$ Rearrange into a standard quadratic equation: $35\lambda^2 - 14\lambda - 21 = 0$ Divide by 7 to simplify: $5\lambda^2 - 2\lambda - 3 = 0$ Reasoning: The given distance provides a second equation that, when combined with the perpendicularity condition, allows us to solve for the parameter $\lambda$.

Step 4: Solve for $\lambda$ and determine the value of $a$ using the condition $a>0$. Factor the quadratic equation $5\lambda^2 - 2\lambda - 3 = 0$: $(5\lambda + 3)(\lambda - 1) = 0$ This yields two possible values for $\lambda$: $\lambda = 1 \quad \text{or} \quad \lambda = -\frac{3}{5}$ Now, we use the condition $a > 0$ with Equation 1 ($a = 7\lambda - 2$):

• Case 1: If $\lambda = 1$ $a = 7(1) - 2 = 5$ Since $a=5 > 0$, this value of $\lambda$ is valid.
• Case 2: If $\lambda = -\frac{3}{5}$ $a = 7\left(-\frac{3}{5}\right) - 2 = -\frac{21}{5} - \frac{10}{5} = -\frac{31}{5}$ Since $a = -\frac{31}{5} < 0$, this value of $\lambda$ is not valid.

Thus, we must have $\lambda = 1$ and $a = 5$. Now, we can find the coordinates of point $P$ and the foot of the perpendicular $R$: Point $P$ is $(a, 4, 2) = (5, 4, 2)$. Foot of perpendicular $R$ (substituting $\lambda = 1$): $R_x = 2(1) - 1 = 1$ $R_y = 3(1) + 3 = 6$ $R_z = -(1) + 1 = 0$ So, the foot of the perpendicular is $R(1, 6, 0)$. Reasoning: The given constraint $a>0$ helps us uniquely identify the correct parameter value from the quadratic solution. Once $\lambda$ and $a$ are known, we can find the exact coordinates of $P$ and $R$.

Step 5: Find the image of point $P$, denoted as $Q(\alpha_1, \alpha_2, \alpha_3)$. The foot of the perpendicular $R$ is the midpoint of the segment connecting point $P$ and its image $Q$. Using the midpoint formula: $R = \left( \frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2}, \frac{z_P + z_Q}{2} \right)$ We have $P(5, 4, 2)$ and $R(1, 6, 0)$. Let $Q(\alpha_1, \alpha_2, \alpha_3)$. For the x-coordinate: $1 = \frac{5 + \alpha_1}{2} \Rightarrow 2 = 5 + \alpha_1 \Rightarrow \alpha_1 = -3$ For the y-coordinate: $6 = \frac{4 + \alpha_2}{2} \Rightarrow 12 = 4 + \alpha_2 \Rightarrow \alpha_2 = 8$ For the z-coordinate: $0 = \frac{2 + \alpha_3}{2} \Rightarrow 0 = 2 + \alpha_3 \Rightarrow \alpha_3 = -2$ So, the image of point $P$ is $Q(-3, 8, -2)$. Reasoning: The geometric property of images in a line (midpoint relationship) is fundamental for finding the coordinates of the image point.

Step 6: Calculate the final expression $a + \sum_{i=1}^{3} \alpha_{i}$. First, calculate the sum of the coordinates of $Q$: $\sum_{i=1}^{3} \alpha_{i} = \alpha_1 + \alpha_2 + \alpha_3 = -3 + 8 + (-2) = 5 - 2 = 3$ Now, add the value of $a$: $a + \sum_{i=1}^{3} \alpha_{i} = 5 + 3 = 8$ Reasoning: This is the final calculation required by the problem statement, combining all previous results.

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Common Mistakes & Tips

• Algebraic Errors: Be very careful with expanding squares and combining like terms, especially with negative signs. A small mistake here can lead to incorrect values for $\lambda$ and $a$.
• Parameter Interpretation: Always use all given conditions (like $a>0$) to correctly choose between multiple solutions for parameters. Skipping this step can lead to an incorrect point or image.
• Conceptual Clarity: Ensure a clear understanding of the difference between a general point on a line, the foot of the perpendicular, and the image of a point. Each has distinct properties and calculation methods.

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Summary

This problem required a systematic application of 3D geometry concepts. We began by parametrizing a general point on the line and forming a vector from the given point $P$ to this general point. The perpendicularity condition between this vector and the line's direction vector provided a relationship between $a$ and the parameter $\lambda$. Using the given perpendicular distance, we formed a quadratic equation in $\lambda$, which, along with the condition $a>0$, allowed us to uniquely determine $\lambda=1$ and $a=5$. With these values, we found the coordinates of $P$ and the foot of the perpendicular $R$. Finally, using the midpoint property for images, we found the coordinates of $Q$ and calculated the required sum $a + \sum \alpha_i = 8$.

The final answer is $\boxed{8}$, which corresponds to option (B).