This problem requires us to find the foot of the perpendicular from a given point to a line. The line itself is defined as the intersection of two planes. This involves a combination of fundamental concepts from 3D Geometry: finding the equation of a line from plane intersections and then applying the condition for perpendicularity to locate the foot of the perpendicular.

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Key Concepts and Formulas:

1. Direction Vector of the Line of Intersection of Two Planes: If two planes are given by their normal vectors $\vec{n_1}$ and $\vec{n_2}$, the line of their intersection L has a direction vector $\vec{d}$ given by the cross product of their normal vectors: $\vec{d} = \vec{n_1} \times \vec{n_2}$
2. Equation of a Line in Parametric Form: A line passing through a point $(x_0, y_0, z_0)$ with direction ratios $(a, b, c)$ can be written in parametric form as: $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} = \lambda$ where $\lambda$ is a scalar parameter. Any point on the line can then be expressed as $(x_0 + a\lambda, y_0 + b\lambda, z_0 + c\lambda)$.
3. Condition for Perpendicular Lines/Vectors: If a line segment AP is perpendicular to a line L (with direction vector $\vec{d_L}$), then the direction vector of AP, $\vec{AP}$, must be perpendicular to $\vec{d_L}$. Their dot product must be zero: $\vec{AP} \cdot \vec{d_L} = 0$

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Step-by-Step Solution

Step 1: Determine the Equation of Line L

The line L is the intersection of the two planes: Plane 1 ($P_1$): $x + 2y + z = 6$ Plane 2 ($P_2$): $y + 2z = 4$

Step 1.1: Find the Direction Vector of L

• Explanation: The line of intersection of two planes is perpendicular to the normal vectors of both planes. Therefore, its direction vector can be found by taking the cross product of the normal vectors of the two planes.
• The normal vector for Plane 1 ($x + 2y + z = 6$) is $\vec{n_1} = (1, 2, 1)$.
• The normal vector for Plane 2 ($0x + y + 2z = 4$) is $\vec{n_2} = (0, 1, 2)$.
• The direction vector $\vec{d_L}$ of line L is given by $\vec{d_L} = \vec{n_1} \times \vec{n_2}$: $\vec{d_L} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{vmatrix}$ $\vec{d_L} = \widehat{i}(2 \times 2 - 1 \times 1) - \widehat{j}(1 \times 2 - 0 \times 1) + \widehat{k}(1 \times 1 - 0 \times 2)$ $\vec{d_L} = (4 - 1)\widehat{i} - (2 - 0)\widehat{j} + (1 - 0)\widehat{k}$ $\vec{d_L} = 3\widehat{i} - 2\widehat{j} + \widehat{k}$
• Thus, the direction ratios (DRs) of line L are $(3, -2, 1)$.

Step 1.2: Find a Point on L

• Explanation: To define a line, we need a point on it in addition to its direction vector. A point on the line of intersection must satisfy both plane equations. We can find such a point by setting one of the coordinates (e.g., $z$) to a convenient value (often 0) and solving the resulting system of two linear equations in two variables.
• Let's set $z = 0$:
  • From Plane 2: $y + 2(0) = 4 \implies y = 4$.
  • Substitute $y=4$ and $z=0$ into Plane 1: $x + 2(4) + 0 = 6 \implies x + 8 = 6 \implies x = -2$.
• So, a point on line L is $A_0(-2, 4, 0)$.

Step 1.3: Write the Parametric Equation of L

• Explanation: Using the point $A_0(-2, 4, 0)$ and the direction ratios $(3, -2, 1)$, we can write the equation of line L in parametric form.
• The equation of line L is: $\frac{x - (-2)}{3} = \frac{y - 4}{-2} = \frac{z - 0}{1} = \lambda$
• Any point P on line L can be represented as $P(\lambda) = (-2 + 3\lambda, 4 - 2\lambda, \lambda)$.

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Step 2: Find the Foot of Perpendicular P

Let the given point be $A = (3, 2, 1)$. Let $P(\alpha, \beta, \gamma)$ be the foot of the perpendicular from A to line L.

Step 2.1: Express Point P in terms of $\lambda$

• Explanation: Since P is the foot of the perpendicular on line L, it must be a point on L. So, its coordinates can be expressed using the parametric form derived in Step 1.3.
• $P(\alpha, \beta, \gamma) = (-2 + 3\lambda, 4 - 2\lambda, \lambda)$.

Step 2.2: Determine the Direction Vector of Line AP

• Explanation: We need the direction vector of the line segment connecting the given point A to the foot of the perpendicular P.
• $\vec{AP} = P - A = ((-2 + 3\lambda) - 3, (4 - 2\lambda) - 2, \lambda - 1)$
• $\vec{AP} = (3\lambda - 5, 2 - 2\lambda, \lambda - 1)$.

Step 2.3: Apply Perpendicularity Condition

• Explanation: The line segment AP is perpendicular to line L. Therefore, their direction vectors must be orthogonal, meaning their dot product is zero. We use the direction vector of L, $\vec{d_L} = (3, -2, 1)$, found in Step 1.1.
• $\vec{AP} \cdot \vec{d_L} = 0$
• $(3\lambda - 5)(3) + (2 - 2\lambda)(-2) + (\lambda - 1)(1) = 0$
• $9\lambda - 15 - 4 + 4\lambda + \lambda - 1 = 0$
• Combine like terms: $(9 + 4 + 1)\lambda + (-15 - 4 - 1) = 0$
• $14\lambda - 20 = 0$
• $14\lambda = 20 \implies \lambda = \frac{20}{14} = \frac{10}{7}$.

Step 2.4: Find the Coordinates of P

• Explanation: Substitute the value of $\lambda$ back into the parametric coordinates of P to find the exact coordinates of the foot of the perpendicular.
• $\alpha = -2 + 3\left(\frac{10}{7}\right) = -2 + \frac{30}{7} = \frac{-14 + 30}{7} = \frac{16}{7}$
• $\beta = 4 - 2\left(\frac{10}{7}\right) = 4 - \frac{20}{7} = \frac{28 - 20}{7} = \frac{8}{7}$
• $\gamma = \frac{10}{7}$
• So, the foot of the perpendicular is $P\left(\frac{16}{7}, \frac{8}{7}, \frac{10}{7}\right)$.

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Step 3: Calculate the Final Value

We need to find the value of $21(\alpha + \beta + \gamma)$.

• First, calculate the sum $\alpha + \beta + \gamma$: $\alpha + \beta + \gamma = \frac{16}{7} + \frac{8}{7} + \frac{10}{7} = \frac{16 + 8 + 10}{7} = \frac{34}{7}$
• Now, multiply by 21: $21(\alpha + \beta + \gamma) = 21 \times \frac{34}{7}$ $21 \times \frac{34}{7} = 3 \times 34 = 102$

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Common Mistakes & Tips:

• Cross Product Accuracy: Be very careful when calculating the cross product of normal vectors. A sign error here will propagate throughout the entire problem.
• Finding a Point on the Line: While setting $z=0$ is common, you can set $x=0$ or $y=0$ as well. Choose the one that simplifies calculations. Ensure the point satisfies both plane equations.
• Algebraic Precision: Solving for $\lambda$ often involves fractions. Maintain precision with calculations involving fractions to avoid errors.
• Conceptual Understanding: Remember why the dot product is used for perpendicularity. This helps in setting up the equation correctly.

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Summary:

This problem demonstrates a complete application of 3D geometry concepts to find the foot of a perpendicular. The key steps involve formulating the line of intersection, using its parametric form to represent the foot of the perpendicular, applying the perpendicularity condition via a dot product to solve for the parameter $\lambda$, and finally calculating the required sum. Our derivation consistently leads to the value 102.

The final answer is $\boxed{102}$ which corresponds to option (A).