Key Concepts and Formulas

1. Direction Vector of a Line Parallel to Two Planes: If a line is parallel to two planes, its direction vector is perpendicular to the normal vectors of both planes. Thus, the direction vector of the line can be found by taking the cross product of the normal vectors of the two planes.
2. Equation of a Line in 3D (Parametric Form): A line passing through a point $(x_0, y_0, z_0)$ with direction vector $\vec{d} = (a, b, c)$ can be represented by the symmetric equations $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} = k$, where $k$ is a scalar parameter. This allows any point on the line to be expressed as $(x_0+ak, y_0+bk, z_0+ck)$.
3. Foot of the Perpendicular: If $P$ is an external point and $F$ is the foot of the perpendicular from $P$ to a line, then the vector $\vec{PF}$ is perpendicular to the direction vector of the line, $\vec{d}$. This implies their dot product is zero: $\vec{PF} \cdot \vec{d} = 0$.

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Step-by-Step Solution

Step 1: Determine the Direction Vector of the Line

The problem states that the line passes through the point $(3, 2, 1)$ and is parallel to two planes: Plane 1: $x+2y+z=0$ Plane 2: $3y-z=3$

• Understanding the condition: A line is parallel to a plane if its direction vector is perpendicular to the plane's normal vector. If the line is parallel to two planes, its direction vector must be perpendicular to the normal vectors of both planes.
• Identify normal vectors:
  • The normal vector for Plane 1 ($x+2y+z=0$) is $\vec{n_1} = (1, 2, 1)$.
  • The normal vector for Plane 2 ($0x+3y-z=3$) is $\vec{n_2} = (0, 3, -1)$.
• Calculate the direction vector: The direction vector of the line, $\vec{b}$, will be perpendicular to both $\vec{n_1}$ and $\vec{n_2}$. Therefore, $\vec{b}$ can be found by taking the cross product of $\vec{n_1}$ and $\vec{n_2}$. $\vec{b} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 0 & 3 & -1 \end{vmatrix}$ $\vec{b} = \hat{i}((2)(-1) - (1)(3)) - \hat{j}((1)(-1) - (1)(0)) + \hat{k}((1)(3) - (2)(0))$ $\vec{b} = \hat{i}(-2 - 3) - \hat{j}(-1 - 0) + \hat{k}(3 - 0)$ $\vec{b} = -5\hat{i} + \hat{j} + 3\hat{k}$ Thus, the direction ratios of the line are $(-5, 1, 3)$.

Step 2: Write the Equation of the Line

We now have a point on the line $(x_0, y_0, z_0) = (3, 2, 1)$ and its direction vector $\vec{b} = (-5, 1, 3)$.

• Formulate the symmetric equation: $\frac{x-3}{-5} = \frac{y-2}{1} = \frac{z-1}{3}$
• Parameterize a general point on the line: Let the common ratio be $k$. This allows us to express the coordinates of any point on the line in terms of a single parameter $k$. $\frac{x-3}{-5} = k \implies x = -5k+3$ $\frac{y-2}{1} = k \implies y = k+2$ $\frac{z-1}{3} = k \implies z = 3k+1$ So, any general point on the line can be represented as $F(-5k+3, k+2, 3k+1)$. This point will be the foot of the perpendicular for a specific value of $k$.

Step 3: Find the Foot of the Perpendicular

Let the given external point be $P(1, 9, 7)$. Let the foot of the perpendicular from $P$ to the line be $F(\alpha, \beta, \gamma)$. From Step 2, we know that $F$ can be expressed as $F(-5k+3, k+2, 3k+1)$.

• Form the vector $\vec{PF}$: We need to find the vector connecting the external point $P$ to the general point $F$ on the line. $\vec{PF} = (x_F - x_P, y_F - y_P, z_F - z_P)$ $\vec{PF} = ((-5k+3) - 1, (k+2) - 9, (3k+1) - 7)$ $\vec{PF} = (-5k+2, k-7, 3k-6)$
• Apply the perpendicularity condition: The vector $\vec{PF}$ must be perpendicular to the direction vector of the line, $\vec{b} = (-5, 1, 3)$. For two vectors to be perpendicular, their dot product must be zero. $\vec{PF} \cdot \vec{b} = 0$ $(-5k+2)(-5) + (k-7)(1) + (3k-6)(3) = 0$ Expand and simplify the equation: $(25k - 10) + (k - 7) + (9k - 18) = 0$ Combine the terms involving $k$ and the constant terms: $(25k + k + 9k) + (-10 - 7 - 18) = 0$ $35k - 35 = 0$ $35k = 35$ $k = 1$
• Determine the coordinates of the foot of the perpendicular: Substitute the value $k=1$ back into the parametric coordinates of $F(-5k+3, k+2, 3k+1)$: $\alpha = -5(1)+3 = -5+3 = -2$ $\beta = 1+2 = 3$ $\gamma = 3(1)+1 = 3+1 = 4$ So, the foot of the perpendicular is $(\alpha, \beta, \gamma) = (-2, 3, 4)$.

Step 4: Calculate $\alpha+\beta+\gamma$

The problem asks for the sum of the coordinates of the foot of the perpendicular: $\alpha+\beta+\gamma = -2 + 3 + 4$ $\alpha+\beta+\gamma = 1 + 4$ $\alpha+\beta+\gamma = 5$

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Common Mistakes & Tips

• Cross Product Errors: Be meticulous when calculating the cross product. A single sign error or incorrect term can lead to an entirely wrong direction vector.
• Incorrect Dot Product Application: Ensure you apply the dot product condition correctly: $\vec{PF} \cdot \vec{b} = 0$, where $\vec{b}$ is the line's direction vector, not the vector connecting two points on the line.
• Parameterization Oversight: Forgetting to parameterize a general point on the line in terms of $k$ or making an error in its algebraic expression is a common pitfall.
• Arithmetic Errors: Double-check all arithmetic, especially when solving for $k$ and substituting it back into the coordinates.

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Summary

To find the foot of the perpendicular from a point to a line, we first determined the line's direction vector by taking the cross product of the normal vectors of the two planes it is parallel to. Next, we wrote the parametric equations of the line using a general parameter $k$. We then formed a vector from the given external point to this general point on the line. By applying the condition that this vector must be perpendicular to the line's direction vector (i.e., their dot product is zero), we solved for the parameter $k$. Finally, substituting the value of $k$ back into the parametric coordinates gave us the coordinates of the foot of the perpendicular, and we summed these coordinates to obtain the final answer.

The final answer is $\boxed{5}$, which corresponds to option (D).