Let the image of the point P(2, -1, 3) in the plane x + 2y - z = 0 be Q. Then the distance of the pl

Q: Wait, why is there a $-2$ in the image formula? Shouldn't it be $-1$?

That is a fantastic question! Think of it this way: the vector from the point $P$ to the plane along the normal is a certain length. To reach the 'foot of the perpendicular' (the point on the plane), you travel that vector once. To reach the 'image' $Q$, you must travel that same vector twice—once to get to the plane, and once more to get to the other side. That is exactly why we use $-2$.

Q: Can I use vector projection instead of the formula? I always forget the formula.

Absolutely! In fact, deriving it via projection is a sign of a true physicist. You can find the normal vector $\vec{n} = (a, b, c)$, project the vector $\vec{PQ}$ onto $\vec{n}$, and then use the displacement vector to find $Q$. The formula is just a shortcut for this exact process. If you ever blank on the formula, just remember: the vector from $P$ to $Q$ must be parallel to the normal vector of the plane.

Q: Is there a trick to simplify the distance calculation at the end? I got $42 / \sqrt{14}$.

You are already there! Don't be intimidated by the radical in the denominator. Instead of just rationalizing, look for factors. Notice that $42 = 3 \times 14$. Since $14 = (\sqrt{14})^2$, the expression becomes $\frac{3 \times (\sqrt{14})^2}{\sqrt{14}}$, which simplifies beautifully to $3\sqrt{14}$. Always look for hidden factors before diving into heavy arithmetic!

Question

Let the image of the point $P(2, -1, 3)$ in the plane $x + 2y - z = 0$ be $Q$. Then the distance of the plane $3x + 2y + z + 29 = 0$ from the point $Q$ is

Options:

(A) $\frac{22\sqrt{2}}{7}$

(B) $\frac{24\sqrt{2}}{7}$

(C) $2\sqrt{14}$

(D) $3\sqrt{14}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a coordinate geometry problem; we are exploring the elegant symmetry of three-dimensional space. In the JEE Advanced, problems involving planes and points are not just about plugging numbers into formulas—they are about visualizing the architecture of space.

Let us break this problem down into its two fundamental acts.

### Act I: The Mirror of the Plane

Imagine you are standing in a room, and there is a mirror placed exactly along the plane defined by $x + 2y - z = 0$. You are holding a point $P(2, -1, 3)$ in your hand. Your goal is to find where the reflection of this point, $Q$, would appear behind the mirror.

This is the essence of the image formula. We are looking for a point $Q(\alpha, \beta, \gamma)$ such that the line segment $PQ$ is perpendicular to the plane, and the plane bisects $PQ$. The formula we use is:

$$\frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = \frac{\gamma - z_1}{c} = -2 \frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}$$

Why the $-2$? I want you to internalize this. If we used $-1$, we would stop exactly at the plane—that is the 'foot of the perpendicular.'

But an image is a reflection; it is as far behind the mirror as the object is in front. Thus, we double the displacement.

Let us substitute our values: $P(2, -1, 3)$ and the plane $x + 2y - z = 0$. Here, $a=1, b=2, c=-1$, and $d=0$. Plugging these in:

$$\frac{\alpha - 2}{1} = \frac{\beta - (-1)}{2} = \frac{\gamma - 3}{-1} = -2 \frac{(1)(2) + (2)(-1) + (-1)(3) + 0}{1^2 + 2^2 + (-1)^2}$$

Look at the numerator of the right-hand side: $2 - 2 - 3 = -3$. The denominator is $1 + 4 + 1 = 6$.

So, the entire right-hand side becomes $-2 	imes (\frac{-3}{6}) = -2 	imes (-0.5) = 1$.

This is the moment where the complexity collapses into simplicity. We now have three simple linear equations:

1. $\alpha - 2 = 1 \Rightarrow \alpha = 3$
2. $\beta + 1 = 2 \Rightarrow \beta = 1$
3. $\gamma - 3 = -1 \Rightarrow \gamma = 2$

We have found our point $Q(3, 1, 2)$. The reflection is complete.

### Act II: The Distance to the Horizon

Now that we have our point $Q(3, 1, 2)$, the problem shifts. We are no longer dealing with reflections. We are now asked to find the perpendicular distance from this point $Q$ to a new, second plane: $3x + 2y + z + 29 = 0$.

Think of this as finding the shortest path from a fixed location to a wall. The formula for the perpendicular distance $d$ from a point $(x_0, y_0, z_0)$ to a plane $ax + by + cz + d = 0$ is a classic, powerful tool in your arsenal:

$$d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$$

Let us carefully substitute our coordinates of $Q(3, 1, 2)$ and the coefficients of the second plane ($a=3, b=2, c=1, d=29$):

$$d = \frac{|3(3) + 2(1) + 1(2) + 29|}{\sqrt{3^2 + 2^2 + 1^2}}$$

Let us calculate the numerator: $9 + 2 + 2 + 29 = 42$. 
Now the denominator: $\sqrt{9 + 4 + 1} = \sqrt{14}$.

So, we have $d = \frac{42}{\sqrt{14}}$.

### The Final Flourish

In the JEE, the final step is often where students lose confidence. Do not panic when you see $\frac{42}{\sqrt{14}}$.

Remember that $42$ is simply $3 	imes 14$. And since $14$ is $(\sqrt{14})^2$, we can rewrite the expression:

$$d = \frac{3 	imes 14}{\sqrt{14}} = 3\sqrt{14}$$

And there it is. The distance is **$3\sqrt{14}$**.

### Reflection for the Student

This problem was a test of two things: your ability to handle the specific mechanics of 3D geometry (the image formula) and your ability to maintain composure through multi-step calculations. You didn't just calculate a number; you navigated through a reflection and then measured a distance.

This is the mindset of an engineer—breaking down a complex, multi-layered problem into manageable, logical steps. Keep this clarity, keep this focus, and you will conquer any problem the exam throws at you.

Animated Solution for Mathematics - Three Dimensional Geometry: Let the image of the point $P (2, - 1, 3)$ in the plane $x + 2 y - z = 0$ be $Q$ . Then the distance of the plane $3 x + 2 y + z + 29 = 0$ from the point $Q$ is

Select Answer:

Visualized Solution

Visualizing Point $P$ and Plane 1

The Image Formula

Substituting Values into the Formula

Simplifying the Right Hand Side

Calculating Coordinates of $Q$

Introducing Plane 2

The Distance Formula

Substituting into Distance Formula

Simplifying the Expression

Final Calculation

Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Act I

Act II

The Final Flourish

Reflection for the Student

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