Key Concepts and Formulas

1. Line Representation: A line in 3D space can be represented parametrically as $R(t) = Q + t\vec{v}$, where $Q(x_0, y_0, z_0)$ is a specific point on the line and $\vec{v} = \langle a, b, c \rangle$ is its direction vector.
2. Direction Vector of Line of Intersection: If a line is given as the intersection of two planes with normal vectors $\vec{n_1}$ and $\vec{n_2}$, its direction vector $\vec{v}$ is parallel to the cross product of the normal vectors: $\vec{v} = \vec{n_1} \times \vec{n_2}$.
3. Foot of the Perpendicular: To find the shortest distance from a point $P$ to a line, we find the foot of the perpendicular, say $R$, on the line. The vector $\vec{PR}$ (from $P$ to $R$) must be perpendicular to the line's direction vector $\vec{v}$. This condition is expressed by their dot product: $\vec{PR} \cdot \vec{v} = 0$.
4. Distance Formula: Once the coordinates of $P(x_P, y_P, z_P)$ and the foot of the perpendicular $R(x_R, y_R, z_R)$ are known, the shortest distance is calculated using the 3D distance formula: $PR = \sqrt{(x_R - x_P)^2 + (y_R - y_P)^2 + (z_R - z_P)^2}$

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Step-by-Step Solution

Step 1: Determine the Direction Ratios and a Point on the Line

The line is given as the intersection of two planes:

• Plane 1 ($S_1$): $3y - 2z - 1 = 0$
• Plane 2 ($S_2$): $3x - z + 4 = 0$

Reasoning: The direction vector of the line of intersection is perpendicular to the normal vectors of both planes. Therefore, we find the direction vector by taking the cross product of their normal vectors.

• The normal vector of Plane 1 (rewritten as $0x + 3y - 2z - 1 = 0$) is $\vec{n_1} = \langle 0, 3, -2 \rangle$.
• The normal vector of Plane 2 (rewritten as $3x + 0y - 1z + 4 = 0$) is $\vec{n_2} = \langle 3, 0, -1 \rangle$.

The direction vector $\vec{d}$ of the line is $\vec{n_1} \times \vec{n_2}$: $\vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 3 & -2 \\ 3 & 0 & -1 \end{vmatrix}$ $\vec{d} = \mathbf{i}((3)(-1) - (-2)(0)) - \mathbf{j}((0)(-1) - (-2)(3)) + \mathbf{k}((0)(0) - (3)(3))$ $\vec{d} = \mathbf{i}(-3 - 0) - \mathbf{j}(0 - (-6)) + \mathbf{k}(0 - 9)$ $\vec{d} = -3\mathbf{i} - 6\mathbf{j} - 9\mathbf{k}$ The direction ratios are $\langle -3, -6, -9 \rangle$. We can simplify these by dividing by a common factor of $-3$ to get $\langle 1, 2, 3 \rangle$. Let's use this simplified direction vector $\vec{v} = \langle 1, 2, 3 \rangle$ for easier calculations.

Next, we need to find a point on the line. Reasoning: To find a point on the line of intersection, we can set one coordinate to zero and solve the resulting system of two linear equations in two variables. Let's set $x=0$:

1. From Plane 2: $3(0) - z + 4 = 0 \Rightarrow -z = -4 \Rightarrow z = 4$
2. From Plane 1: $3y - 2(4) - 1 = 0 \Rightarrow 3y - 8 - 1 = 0 \Rightarrow 3y = 9 \Rightarrow y = 3$ So, a point $Q$ on the line is $(0, 3, 4)$.

The parametric form of any point $R$ on the line, using $Q(0,3,4)$ and direction vector $\vec{v} = \langle 1, 2, 3 \rangle$, is: $R(\lambda) = (0 + 1\lambda, 3 + 2\lambda, 4 + 3\lambda) = (\lambda, 3 + 2\lambda, 4 + 3\lambda)$

Step 2: Find the Foot of the Perpendicular

Let the given point be $P(2, -1, 6)$. Let $R(\lambda, 3 + 2\lambda, 4 + 3\lambda)$ be the foot of the perpendicular from $P$ to the line.

Reasoning: The vector $\vec{PR}$ must be perpendicular to the line's direction vector $\vec{v}$. Their dot product must be zero.

First, form the vector $\vec{PR}$: $\vec{PR} = (x_R - x_P)\mathbf{i} + (y_R - y_P)\mathbf{j} + (z_R - z_P)\mathbf{k}$ $\vec{PR} = (\lambda - 2)\mathbf{i} + ((3 + 2\lambda) - (-1))\mathbf{j} + ((4 + 3\lambda) - 6)\mathbf{k}$ $\vec{PR} = (\lambda - 2)\mathbf{i} + (4 + 2\lambda)\mathbf{j} + (3\lambda - 2)\mathbf{k}$

Now, apply the perpendicularity condition $\vec{PR} \cdot \vec{v} = 0$: $(\lambda - 2)(1) + (4 + 2\lambda)(2) + (3\lambda - 2)(3) = 0$ $\lambda - 2 + 8 + 4\lambda + 9\lambda - 6 = 0$ Combine like terms: $(\lambda + 4\lambda + 9\lambda) + (-2 + 8 - 6) = 0$ $14\lambda + 0 = 0$ $\lambda = 0$

Reasoning: The value $\lambda = 0$ indicates that the point $Q(0,3,4)$ we initially found on the line is, in fact, the foot of the perpendicular from $P$ to the line.

Substitute $\lambda = 0$ back into the parametric coordinates of $R$ to find the foot of the perpendicular: $x_R = 0$ $y_R = 3 + 2(0) = 3$ $z_R = 4 + 3(0) = 4$ So, the foot of the perpendicular is $R(0, 3, 4)$.

Step 3: Calculate the Distance PR

Now we have the given point $P(2, -1, 6)$ and the foot of the perpendicular $R(0, 3, 4)$. Reasoning: The shortest distance is simply the Euclidean distance between these two points.

The shortest distance $PR$ is calculated using the distance formula: $PR = \sqrt{(x_R - x_P)^2 + (y_R - y_P)^2 + (z_R - z_P)^2}$ $PR = \sqrt{(0 - 2)^2 + (3 - (-1))^2 + (4 - 6)^2}$ $PR = \sqrt{(-2)^2 + (4)^2 + (-2)^2}$ $PR = \sqrt{4 + 16 + 6}$ $PR = \sqrt{26}$

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Common Mistakes & Tips

• Cross Product Calculation: Be meticulous with signs and order of operations when computing the cross product of normal vectors, especially for the $\mathbf{j}$ component.
• Finding a Point on the Line: Choose values for coordinates that simplify the plane equations (e.g., $x=0$, $y=0$, or $z=0$) to avoid complex fractions when finding a point on the line.
• Dot Product for Perpendicularity: Remember that the dot product of two perpendicular vectors is zero. This is a crucial step for finding the parameter $\lambda$.
• Arithmetic: Double-check all additions, subtractions, multiplications, and squaring, particularly with negative numbers, to prevent calculation errors.

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Summary

To find the distance from the point $(2, -1, 6)$ to the given line, we first determined the line's direction vector $\langle 1, 2, 3 \rangle$ and a point on the line $(0, 3, 4)$. Then, we set up the parametric form for any point on the line and used the condition that the vector from the given point to the foot of the perpendicular on the line must be orthogonal to the line's direction vector. This led to finding the foot of the perpendicular at $(0, 3, 4)$. Finally, we calculated the distance between the given point and the foot of the perpendicular using the 3D distance formula, yielding $\sqrt{26}$.

The final answer is $\boxed{\sqrt{26}}$, which corresponds to option (A).