Here's a detailed, step-by-step solution for the given problem, adhering to the requested structure and aiming for clarity and educational value.

1. Key Concepts and Formulas

   • Equation of a Line in Symmetric Form: A line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ can be written as $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.
   • General Point on a Line: To represent any point on a line given in symmetric form, we introduce a parameter (e.g., $\lambda$) by setting each ratio equal to this parameter. This allows us to express the coordinates $(x, y, z)$ of any point on the line in terms of $\lambda$.
   • Equation of a Plane: A linear equation in $x, y, z$ of the form $Ax + By + Cz = D$ represents a plane.
   • Intersection of a Line and a Plane: The point where a line meets a plane must satisfy both the equation of the line and the equation of the plane. By substituting the parameterized coordinates of a general point on the line into the plane's equation, we can find the specific value of the parameter that corresponds to the intersection point.
   • Distance Formula in 3D: The distance of a point $P(x, y, z)$ from the origin $O(0, 0, 0)$ is given by the formula $OP = \sqrt{x^2 + y^2 + z^2}$.

2. Step-by-Step Solution

   Step 1: Parameterize the Line

   The given equation of the line is: $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ To find a general point on this line, we set each ratio equal to a parameter, say $\lambda$. This allows us to express the coordinates $(x, y, z)$ of any point on the line in terms of $\lambda$. $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4} = \lambda$ From this, we can solve for $x, y,$ and $z$ in terms of $\lambda$:

   • From $\frac{x - 1}{2} = \lambda \Rightarrow x - 1 = 2\lambda \Rightarrow x = 2\lambda + 1$
   • From $\frac{y + 1}{3} = \lambda \Rightarrow y + 1 = 3\lambda \Rightarrow y = 3\lambda - 1$
   • From $\frac{z - 2}{4} = \lambda \Rightarrow z - 2 = 4\lambda \Rightarrow z = 4\lambda + 2$ Thus, any point P on the line can be represented by the coordinates $(2\lambda + 1, 3\lambda - 1, 4\lambda + 2)$.

   Step 2: Substitute Point P into the Plane Equation

   The line meets the plane $x + 2y + 3z = 15$ at point P. This means the coordinates of point P must satisfy the equation of the plane. We substitute the parameterized coordinates of P into the plane equation: $(2\lambda + 1) + 2(3\lambda - 1) + 3(4\lambda + 2) = 15$

   Step 3: Solve for the Parameter $\lambda$

   Now, we simplify and solve the equation for $\lambda$: $2\lambda + 1 + 6\lambda - 2 + 12\lambda + 6 = 15$ Combine the terms involving $\lambda$ and the constant terms: $(2\lambda + 6\lambda + 12\lambda) + (1 - 2 + 6) = 15$ $20\lambda + 5 = 15$ Subtract 5 from both sides: $20\lambda = 15 - 5$ $20\lambda = 10$ Divide by 20 to find $\lambda$: $\lambda = \frac{10}{20}$ $\lambda = \frac{1}{2}$

   Step 4: Determine the Coordinates of Point P

   Now that we have the value of $\lambda = \frac{1}{2}$, we substitute it back into the parameterized coordinates of P from Step 1:

   • $x = 2\left(\frac{1}{2}\right) + 1 = 1 + 1 = 2$
   • $y = 3\left(\frac{1}{2}\right) - 1 = \frac{3}{2} - 1 = \frac{3 - 2}{2} = \frac{1}{2}$
   • $z = 4\left(\frac{1}{2}\right) + 2 = 2 + 2 = 4$ So, the coordinates of the intersection point P are $\left(2, \frac{1}{2}, 4\right)$.

   Step 5: Calculate the Distance of P from the Origin

   We need to find the distance of point P $\left(2, \frac{1}{2}, 4\right)$ from the origin $O(0,0,0)$. Using the 3D distance formula from the origin $OP = \sqrt{x^2 + y^2 + z^2}$: $OP = \sqrt{(2)^2 + \left(\frac{1}{2}\right)^2 + (4)^2}$ $OP = \sqrt{4 + \frac{1}{4} + 16}$ Combine the integer terms: $OP = \sqrt{20 + \frac{1}{4}}$ To add the terms, find a common denominator: $OP = \sqrt{\frac{80}{4} + \frac{1}{4}}$ $OP = \sqrt{\frac{81}{4}}$ Take the square root of the numerator and the denominator: $OP = \frac{\sqrt{81}}{\sqrt{4}}$ $OP = \frac{9}{2}$

3. Common Mistakes & Tips

   • Algebraic Precision: Errors in simplifying the equation for $\lambda$ or substituting its value can lead to incorrect coordinates for P. Double-check all arithmetic.
   • Parameterization: Ensure each coordinate ($x, y, z$) is correctly expressed in terms of $\lambda$. A common mistake is to incorrectly isolate the variable.
   • Distance Formula: Remember the correct 3D distance formula. For distance from the origin, it's a direct application of $\sqrt{x^2+y^2+z^2}$.
   • Verification: After finding P, a quick check by substituting its coordinates back into both the line and plane equations can help catch errors.

4. Summary

   To find the distance of the intersection point P from the origin, we first parameterized the given line to represent any point on it. We then substituted these parameterized coordinates into the plane equation to solve for the specific parameter value $\lambda$ that corresponds to the intersection point. Using this value of $\lambda$, we determined the exact coordinates of P. Finally, we applied the 3D distance formula to calculate the distance of P from the origin. The calculated distance is $\frac{9}{2}$.

5. Final Answer

   The distance of P from the origin is $\frac{9}{2}$. This corresponds to option (C).

   The final answer is $\boxed{\frac{9}{2}}$