1. Key Concepts and Formulas

This problem integrates concepts from 3D Geometry and basic Trigonometry. We will utilize the following fundamental formulas and properties:

• Distance from a Point to a Plane: The perpendicular distance $d$ from a point $P(x_0, y_0, z_0)$ to a plane $Ax+By+Cz+D=0$ is given by the formula: $d = \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$
• Trigonometric Ratios in a Right-Angled Triangle: For a right-angled triangle, if $\theta$ is one of the acute angles, the tangent of the angle is defined as the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle: $\tan \theta = \frac{\text{Opposite Side}}{\text{Adjacent Side}}$
• Area of a Right-Angled Triangle: The area of a right-angled triangle is half the product of the lengths of its two perpendicular sides (legs): $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}$

2. Step-by-Step Solution

Step 1: Calculate the length of $PQ$, which is the perpendicular distance from point $P$ to the plane.

• Understanding the setup: We are given the point $P(1,2,3)$ and the equation of the plane $x+2y+z=14$. The point $Q$ is defined as the foot of the perpendicular drawn from $P$ to this plane. This means the length of the segment $PQ$ is precisely the perpendicular distance from point $P$ to the given plane.

• Applying the distance formula:

  • The coordinates of the point are $(x_0, y_0, z_0) = (1,2,3)$.
  • The equation of the plane is $x+2y+z=14$, which can be rewritten in the standard form $Ax+By+Cz+D=0$ as $x+2y+z-14=0$.
  • From the plane equation, we identify the coefficients: $A=1, B=2, C=1, D=-14$.

  Substitute these values into the distance formula: $PQ = \frac{|(1)(1) + (2)(2) + (1)(3) - 14|}{\sqrt{1^2 + 2^2 + 1^2}}$ $PQ = \frac{|1 + 4 + 3 - 14|}{\sqrt{1 + 4 + 1}}$ $PQ = \frac{|8 - 14|}{\sqrt{6}}$ $PQ = \frac{|-6|}{\sqrt{6}}$ $PQ = \frac{6}{\sqrt{6}}$ To rationalize the denominator, we multiply the numerator and denominator by $\sqrt{6}$: $PQ = \frac{6\sqrt{6}}{6}$ $PQ = \sqrt{6}$ So, the length of the perpendicular segment $PQ$ is $\sqrt{6}$ units.

Step 2: Determine the length of $QR$ using trigonometry.

• Understanding the geometry: Since $Q$ is the foot of the perpendicular from $P$ to the plane, the line segment $PQ$ is perpendicular to the plane. Any line segment lying in the plane and passing through $Q$ will be perpendicular to $PQ$. Since $R$ is a point on the plane, the line segment $QR$ lies within the plane. Therefore, $PQ$ is perpendicular to $QR$, which means $\triangle PQR$ is a right-angled triangle with the right angle at $Q$ ($\angle PQR = 90^\circ$).
• We are given that $\angle PRQ = 60^\circ$.
• In the right-angled triangle $\triangle PQR$:
  • The side $PQ$ is opposite to $\angle PRQ$.
  • The side $QR$ is adjacent to $\angle PRQ$.
• Applying the trigonometric ratio: We use the tangent function, which relates the opposite side, adjacent side, and the angle: $\tan(\angle PRQ) = \frac{\text{Opposite Side}}{\text{Adjacent Side}} = \frac{PQ}{QR}$ Substitute the known values: $PQ = \sqrt{6}$ (from Step 1) and $\angle PRQ = 60^\circ$. $\tan(60^\circ) = \frac{\sqrt{6}}{QR}$ We know that the exact value of $\tan(60^\circ)$ is $\sqrt{3}$. $\sqrt{3} = \frac{\sqrt{6}}{QR}$ Now, we solve for $QR$: $QR = \frac{\sqrt{6}}{\sqrt{3}}$ $QR = \sqrt{\frac{6}{3}}$ $QR = \sqrt{2}$ Thus, the length of the segment $QR$ is $\sqrt{2}$ units.

Step 3: Calculate the area of $\triangle PQR$.

• Understanding the formula: As established in Step 2, $\triangle PQR$ is a right-angled triangle with the right angle at $Q$. Its area can be calculated as half the product of the lengths of its two perpendicular sides, $PQ$ and $QR$.
• Applying the area formula: $\text{Area}(\triangle PQR) = \frac{1}{2} \times PQ \times QR$ Substitute the calculated values: $PQ = \sqrt{6}$ and $QR = \sqrt{2}$. $\text{Area}(\triangle PQR) = \frac{1}{2} \times \sqrt{6} \times \sqrt{2}$ $\text{Area}(\triangle PQR) = \frac{1}{2} \times \sqrt{6 \times 2}$ $\text{Area}(\triangle PQR) = \frac{1}{2} \times \sqrt{12}$ To simplify $\sqrt{12}$, we factor out the perfect square: $\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}$. $\text{Area}(\triangle PQR) = \frac{1}{2} \times 2\sqrt{3}$ $\text{Area}(\triangle PQR) = \sqrt{3}$ The area of $\triangle PQR$ is $\sqrt{3}$ square units.

3. Common Mistakes & Tips

• Absolute Value: Always remember to use the absolute value in the numerator of the distance formula from a point to a plane, as distance must be a non-negative quantity.
• Identifying the Right Angle: Carefully interpret the problem statement to correctly identify where the right angle of the triangle is. "Foot of perpendicular from $P$ to the plane" implies that $PQ$ is perpendicular to any line in the plane passing through $Q$, making $\angle PQR = 90^\circ$.
• Standard Trigonometric Values: Be proficient with the values of trigonometric functions for common angles (e.g., $30^\circ, 45^\circ, 60^\circ$).
• Simplifying Radicals: Always simplify square roots to their simplest form to match the given options or for a cleaner final answer (e.g., $\sqrt{12} = 2\sqrt{3}$).

4. Summary

This problem effectively demonstrates the application of fundamental concepts in 3D geometry and trigonometry. We began by calculating the perpendicular distance from the given point to the plane, which gave us the length of $PQ$. Recognizing that $PQ$ is perpendicular to any line in the plane through $Q$, we established that $\triangle PQR$ is a right-angled triangle. Using the given angle $\angle PRQ = 60^\circ$ and the tangent trigonometric ratio, we then found the length of the side $QR$. Finally, we calculated the area of the right-angled triangle using the formula $\frac{1}{2} \times PQ \times QR$.

5. Final Answer

The final answer is $\boxed{\sqrt{3}}$, which corresponds to option (A).