Key Concepts and Formulas

• Equation of a Line in 3D (Symmetric Form): A line passing through a point $(x_1, y_1, z_1)$ and having direction ratios (DRs) $\langle a, b, c \rangle$ can be represented as: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$
• Parametric Form of a Line: Any general point on a line defined by the symmetric form can be expressed parametrically by setting the ratios equal to a parameter, say $\lambda$: $x = x_1 + a\lambda, \quad y = y_1 + b\lambda, \quad z = z_1 + c\lambda$
• Distance between Two Points in 3D: The Euclidean distance $D$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by: $D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$ The square of the distance is $D^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$.

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Step-by-Step Solution

This problem asks for the distance of a point $P$ from a plane, measured along a specific line direction. This means we need to find the point of intersection $Q$ of the line (passing through $P$ and parallel to the given direction) with the plane. The required "distance" will then be interpreted as the square of the Euclidean distance between points $P$ and $Q$, based on the provided options and correct answer.

Step 1: Determine the Direction Ratios (DRs) of the Line Along Which the Distance is Measured

The distance is to be measured along a line parallel to the given line: ${{2 - x} \over 2} = {{y - 3} \over 2} = {{z + 1} \over 1}$. To identify the direction ratios, we must first convert this equation into the standard symmetric form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$. The term $\frac{2 - x}{2}$ can be rewritten as $\frac{-(x - 2)}{2} = \frac{x - 2}{-2}$. The other terms are already in standard form: $\frac{y - 3}{2}$ and $\frac{z - (-1)}{1}$. So, the standard form of the line equation is: $\frac{x - 2}{-2} = \frac{y - 3}{2} = \frac{z - (-1)}{1}$ From this, the direction ratios of the line are $\langle -2, 2, 1 \rangle$. Since the line $PQ$ (along which we need to find the distance) is parallel to this given line, its direction ratios will be the same. Therefore, the direction ratios of line $PQ$ are $\langle -2, 2, 1 \rangle$.

Why this step? To define the unique line $PQ$, we need a point it passes through (given as $P$) and its direction. The given line provides this direction.

Step 2: Form the Equation of Line PQ

We are given the point $P(3, 2, -1)$ and we found the direction ratios $\langle -2, 2, 1 \rangle$ for the line $PQ$. Using the standard symmetric form for a line: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ Substituting the coordinates of $P(3, 2, -1)$ for $(x_1, y_1, z_1)$ and $\langle -2, 2, 1 \rangle$ for $\langle a, b, c \rangle$: $\frac{x - 3}{-2} = \frac{y - 2}{2} = \frac{z - (-1)}{1}$ $\Rightarrow \frac{x - 3}{-2} = \frac{y - 2}{2} = \frac{z + 1}{1}$ This is the equation of the line $PQ$.

Why this step? We need to define the line $PQ$ explicitly so that we can find any point on it, particularly the point where it intersects the given plane.

Step 3: Find the Point of Intersection $Q$ of Line PQ with the Plane

To find the coordinates of any general point on the line $PQ$, we set the ratios from the line equation equal to a parameter $\lambda$: $\frac{x - 3}{-2} = \frac{y - 2}{2} = \frac{z + 1}{1} = \lambda$ From this, we can express the coordinates $(x, y, z)$ of any point on the line in terms of $\lambda$: $x - 3 = -2\lambda \Rightarrow x = 3 - 2\lambda$ $y - 2 = 2\lambda \Rightarrow y = 2 + 2\lambda$ $z + 1 = \lambda \Rightarrow z = -1 + \lambda$ So, any point on the line $PQ$ can be represented as $Q(3 - 2\lambda, 2 + 2\lambda, -1 + \lambda)$.

The point $Q$ is where the line $PQ$ intersects the given plane $3x - y + 4z + 1 = 0$. This means the coordinates of $Q$ must satisfy the plane's equation. Substitute the parametric expressions for $x, y, z$ into the plane equation: $3(3 - 2\lambda) - (2 + 2\lambda) + 4(-1 + \lambda) + 1 = 0$ Now, solve this linear equation for $\lambda$: $9 - 6\lambda - 2 - 2\lambda - 4 + 4\lambda + 1 = 0$ Combine the constant terms and the $\lambda$ terms: $(9 - 2 - 4 + 1) + (-6\lambda - 2\lambda + 4\lambda) = 0$ $4 - 4\lambda = 0$ $4\lambda = 4$ $\lambda = 1$

Now that we have the value of $\lambda$, we can find the exact coordinates of point $Q$ by substituting $\lambda = 1$ back into the parametric equations: $x_Q = 3 - 2(1) = 3 - 2 = 1$ $y_Q = 2 + 2(1) = 2 + 2 = 4$ $z_Q = -1 + 1 = 0$ So, the point of intersection is $Q(1, 4, 0)$.

Why this step? Finding $\lambda$ allows us to pinpoint the exact location of the intersection point $Q$. This point $Q$ is crucial because the "distance" we need to calculate is related to the segment $PQ$.

Step 4: Calculate the Squared Distance Between Points P and Q

We have the coordinates of point $P(3, 2, -1)$ and point $Q(1, 4, 0)$. The square of the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by: $D^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$ Substitute the coordinates of $P$ and $Q$: $PQ^2 = (1 - 3)^2 + (4 - 2)^2 + (0 - (-1))^2$ $PQ^2 = (-2)^2 + (2)^2 + (1)^2$ $PQ^2 = 4 + 4 + 1$ $PQ^2 = 9$ Therefore, the square of the distance of the point $P$ from the plane along the given line is $9$ units. Given the options, it is implied that the question is asking for the square of the distance.

Why this step? This is the final calculation to arrive at the quantity asked for, which, based on the provided correct answer, is the square of the Euclidean distance $PQ$.

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Common Mistakes & Tips

• Standard Form for Direction Ratios: Always ensure the line equation is in the standard symmetric form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ before extracting direction ratios. Pay close attention to terms like $(2-x)$ which should be rewritten as $-(x-2)$ to correctly identify the direction ratio as $-2$, not $2$.
• Parametric Form is Essential: Using the parametric form of the line ($x = x_1 + a\lambda$, etc.) is the most reliable method for finding the intersection point with a plane.
• Algebraic Accuracy: Be meticulous with signs and calculations when substituting the parametric form into the plane equation and solving for $\lambda$. Errors here are common and propagate through the rest of the solution.

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Summary

To find the "distance" of a point $P$ from a plane along a specified line direction, we first determine the direction ratios of the line. Then, we construct the equation of the line passing through $P$ with these direction ratios and express it in parametric form. We find the intersection point $Q$ of this line with the plane by substituting the parametric coordinates into the plane's equation to solve for the parameter $\lambda$. Finally, we calculate the square of the Euclidean distance between the initial point $P$ and the intersection point $Q$. In this case, the square of the distance $PQ$ was found to be $9$.

The final answer is $\boxed{9}$ which corresponds to option (A).