Key Concepts and Formulas

This problem requires us to perform two main operations in 3D geometry:

• Finding the Image of a Point in a Plane: The image of a point $P(x_1, y_1, z_1)$ in a plane $Ax+By+Cz+D=0$ is another point $Q(x', y', z')$ such that the plane acts as the perpendicular bisector of the line segment $PQ$. The coordinates of $Q$ can be found using the direct formula: $\frac{x'-x_1}{A} = \frac{y'-y_1}{B} = \frac{z'-z_1}{C} = -2 \frac{Ax_1+By_1+Cz_1+D}{A^2+B^2+C^2}$ This formula is derived by considering the line passing through $P$ and perpendicular to the plane. The foot of the perpendicular, say $M$, lies on the plane and is the midpoint of $PQ$. The factor of $-2$ ensures that we find the coordinates of $Q$, which is twice the distance from $P$ to $M$ along the perpendicular line.

• Finding the Perpendicular Distance of a Point from a Plane: The perpendicular distance $d$ of a point $Q(x_0, y_0, z_0)$ from a plane $Ax+By+Cz+D=0$ is given by the formula: $d = \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$ This formula directly calculates the shortest distance from the given point to the plane. The numerator evaluates the plane equation at the point's coordinates (its absolute value indicates how far the point is from satisfying the plane equation), and the denominator normalizes it by the magnitude of the normal vector of the plane.

Step-by-Step Solution

Step 1: Find the coordinates of the image point Q.

We need to find the image $Q(x', y', z')$ of the point $P(2,-1,3)$ in the plane $x+2y+z=0$.

• Identify the given point and plane parameters:

  • Point $P(x_1, y_1, z_1) = (2, -1, 3)$
  • Plane equation: $x+2y+z=0$. Comparing this with $Ax+By+Cz+D=0$, we have $A=1, B=2, C=1, D=0$.

• Apply the image formula: Substitute these values into the formula for finding the image of a point in a plane: $\frac{x'-x_1}{A} = \frac{y'-y_1}{B} = \frac{z'-z_1}{C} = -2 \frac{Ax_1+By_1+Cz_1+D}{A^2+B^2+C^2}$ $\frac{x'-2}{1} = \frac{y'-(-1)}{2} = \frac{z'-3}{1} = -2 \frac{(1)(2) + (2)(-1) + (1)(3) + 0}{1^2+2^2+1^2}$

• Calculate the constant value (right-hand side): First, evaluate the numerator of the fraction: $(1)(2) + (2)(-1) + (1)(3) + 0 = 2 - 2 + 3 = 3$. Next, evaluate the denominator: $1^2+2^2+1^2 = 1+4+1 = 6$. So, the constant value is: $-2 \frac{3}{6} = -2 \left(\frac{1}{2}\right) = -1$ Let this constant be $\lambda = -1$.

• Solve for the coordinates of Q($x', y', z'$): Now, equate each part of the formula to $\lambda = -1$:

  1. $\frac{x'-2}{1} = -1 \implies x'-2 = -1 \implies x' = 1$
  2. $\frac{y'+1}{2} = -1 \implies y'+1 = -2 \implies y' = -3$
  3. $\frac{z'-3}{1} = -1 \implies z'-3 = -1 \implies z' = 2$

• Coordinates of Q: The image point $Q$ is $(1, -3, 2)$.

Step 2: Find the distance of point Q from the second plane.

Now we need to find the perpendicular distance of the point $Q(1,-3,2)$ from the plane $3x+2y+z+29=0$.

• Identify the point and the second plane parameters:

  • Point $Q(x_0, y_0, z_0) = (1, -3, 2)$
  • Plane equation: $3x+2y+z+29=0$. Comparing this with $Ax+By+Cz+D=0$, we have $A=3, B=2, C=1, D=29$.

• Apply the distance formula: Substitute these values into the distance formula: $d = \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$ $d = \frac{|(3)(1) + (2)(-3) + (1)(2) + 29|}{\sqrt{3^2+2^2+1^2}}$

• Calculate the numerator and denominator:

  • Numerator: Calculate the value inside the absolute sign: $3 - 6 + 2 + 29 = -3 + 2 + 29 = -1 + 29 = 28$. So, the numerator is $|28| = 28$.
  • Denominator: Calculate the square root of the sum of squares of coefficients: $\sqrt{9 + 4 + 1} = \sqrt{14}$.

• Calculate the distance: $d = \frac{28}{\sqrt{14}}$ To rationalize the denominator, multiply both the numerator and denominator by $\sqrt{14}$: $d = \frac{28 \sqrt{14}}{14} = 2\sqrt{14}$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with negative signs, especially when substituting coordinates or coefficients into the formulas. A common mistake is forgetting the negative sign in front of $C$ if the plane equation has $-z$.
• Image vs. Foot of Perpendicular: Remember the factor of $-2$ in the image formula. If you use $-1$ instead, you'll find the foot of the perpendicular, not the image.
• Absolute Value: Always include the absolute value in the numerator for distance calculations, as distance must be non-negative.
• Rationalizing Denominators: Simplify expressions by rationalizing the denominator. This often helps in matching the final answer with the given options.

Summary

We began by identifying the given point $P(2,-1,3)$ and the first plane $x+2y+z=0$. Using the formula for the image of a point in a plane, we calculated the coordinates of the image point $Q$ as $(1,-3,2)$. Subsequently, we used the distance formula to find the perpendicular distance of this point $Q$ from the second plane $3x+2y+z+29=0$. The calculated distance was $2\sqrt{14}$.

Final Answer

The distance of the plane $3x+2y+z+29=0$ from the point $Q$ is $2\sqrt{14}$, which corresponds to option (A).

The final answer is \boxed{\text{2\sqrt{14}}}.