Key Concepts and Formulas:

This problem asks for the distance of a point from a plane, not along the normal to the plane (which would be the shortest, perpendicular distance), but measured parallel to a given line. This means we are looking for the length of a line segment that starts at the given point, is parallel to the specified direction, and ends on the plane.

1. Parametric Equation of a Line: A line passing through a point $A(x_1, y_1, z_1)$ and parallel to a vector $\vec{d} = (a, b, c)$ can be represented parametrically as: $x = x_1 + a\lambda$ $y = y_1 + b\lambda$ $z = z_1 + c\lambda$ where $\lambda$ is a scalar parameter. Each value of $\lambda$ corresponds to a unique point on the line.
2. Point of Intersection: To find where this line intersects a plane, we substitute the parametric equations of the line into the equation of the plane. Solving for $\lambda$ gives the specific parameter value for the intersection point.
3. Distance Formula: The distance between the initial point $A$ and the intersection point $P$ can be found using the distance formula. Alternatively, if the line segment $AP$ is represented by $\lambda \vec{d}$, the distance $AP = |\lambda| \cdot |\vec{d}|$, where $|\vec{d}|$ is the magnitude of the direction vector $\vec{d}$. The magnitude of a vector $\vec{d} = (a,b,c)$ is $|\vec{d}| = \sqrt{a^2 + b^2 + c^2}$.

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Step-by-Step Solution:

1. Identify the Given Information: We are given:

• The initial point $A = (1, -2, 3)$.
• The equation of the plane $\Pi: x - y + z = 5$.
• The direction ratios of the line parallel to which the distance is measured: $(2, 3, -6)$. This gives us the direction vector $\vec{d} = (2, 3, -6)$.

2. Formulate the Parametric Equation of the Line: We need to define a line that passes through point $A(1, -2, 3)$ and is parallel to the direction vector $\vec{d} = (2, 3, -6)$. Using the parametric form of a line: $L: \begin{cases} x = 1 + 2\lambda \\ y = -2 + 3\lambda \\ z = 3 - 6\lambda \end{cases}$ Explanation: Every point on this line can be expressed in terms of the parameter $\lambda$. When $\lambda = 0$, the point is $A(1, -2, 3)$. As $\lambda$ varies, we trace points along the line starting from $A$ in the direction of $\vec{d}$.

3. Find the Point of Intersection with the Plane: Let $P$ be the point where the line $L$ intersects the plane $\Pi$. Since $P$ lies on the plane, its coordinates must satisfy the plane's equation $x - y + z = 5$. Substitute the parametric expressions for $x, y, z$ from the line equation into the plane equation: $(1 + 2\lambda) - (-2 + 3\lambda) + (3 - 6\lambda) = 5$ Explanation: We are looking for the specific value of $\lambda$ that corresponds to the point $P$, which is common to both the line and the plane.

4. Solve for the Parameter $\lambda$: Now, simplify and solve the equation for $\lambda$: $1 + 2\lambda + 2 - 3\lambda + 3 - 6\lambda = 5$ Combine the constant terms and the $\lambda$ terms: $(1 + 2 + 3) + (2\lambda - 3\lambda - 6\lambda) = 5$ $6 - 7\lambda = 5$ Subtract 6 from both sides: $-7\lambda = 5 - 6$ $-7\lambda = -1$ Divide by -7: $\lambda = \frac{-1}{-7} = \frac{1}{7}$ Explanation: This value of $\lambda$ tells us how far along the direction vector $\vec{d}$ we need to travel from point $A$ to reach the plane.

5. Calculate the Distance AP: The distance $AP$ is the magnitude of the vector $\lambda \vec{d}$. First, calculate the magnitude of the direction vector $\vec{d} = (2, 3, -6)$: $|\vec{d}| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ Now, multiply this magnitude by the absolute value of $\lambda$: $AP = |\lambda| \cdot |\vec{d}| = \left|\frac{1}{7}\right| \cdot 7 = \frac{1}{7} \cdot 7 = 1$ Explanation: This method efficiently calculates the distance without needing to find the explicit coordinates of the intersection point $P$ and then using the full distance formula. The distance is the length of the line segment from $A$ to $P$.

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Common Mistakes & Tips:

• Don't Confuse with Perpendicular Distance: This problem explicitly asks for the distance "measured parallel to a line," which is different from the shortest (perpendicular) distance from a point to a plane. Do not use the perpendicular distance formula $\frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
• Parametric Form is Essential: Correctly setting up the parametric equations of the line is the foundation of this problem. Ensure the initial point and direction ratios are used correctly.
• Efficient Distance Calculation: Once $\lambda$ and $|\vec{d}|$ are known, calculating the distance as $|\lambda| \cdot |\vec{d}|$ is often quicker and less prone to arithmetic errors with fractions than finding the intersection point's coordinates and then applying the full distance formula.

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Summary:

To find the distance of a point from a plane measured parallel to a given line, we first construct the parametric equation of a line passing through the given point and parallel to the given direction. We then find the point of intersection of this line with the plane by substituting the line's coordinates into the plane's equation and solving for the parameter $\lambda$. Finally, the distance between the initial point and this intersection point is calculated, most efficiently by multiplying the absolute value of the parameter $\lambda$ by the magnitude of the direction vector. Following these steps, the distance is found to be 1.

The final answer is $\boxed{\text{1}}$, which corresponds to option (D).