Let Q and R be two points on the line (x+1)/2 = (y+2)/3 = (z-1)/2 at a distance √26 from the point P

Q: Why do we drop a perpendicular? Can't we just use the distance formula for $Q$ and $R$ directly?

You absolutely could! You could set the distance from $P$ to a general point on the line equal to $\sqrt{26}$ and solve the resulting quadratic equation for $\lambda$. However, that path is algebraically heavy and prone to calculation errors. Dropping the perpendicular is the 'geometric shortcut'—it breaks the problem into smaller, manageable, and more elegant pieces.

Q: Why is the dot product of $\vec{PM}$ and the line's direction vector zero?

This is the heart of 3D geometry! If a line segment is perpendicular to a line, then the vector representing that segment must be orthogonal to the line's direction vector. The dot product of two orthogonal vectors is always zero. It is the most powerful tool in your 3D geometry toolkit.

Q: Why do we multiply $QM$ by 2 at the end?

Because $\triangle PQR$ is isosceles with $PQ = PR$. In any isosceles triangle, the altitude drawn from the vertex to the base bisects the base. Therefore, the foot of the perpendicular $M$ is the midpoint of $QR$. So, $QR = 2 \times QM$.

Question

Let $Q$ and $R$ be two points on the line $\frac{x+1}{2} = \frac{y+2}{3} = \frac{z-1}{2}$ at a distance $\sqrt{26}$ from the point $P(4, 2, 7)$. Then the square of the area of the triangle $PQR$ is ______.

Sigma Percentile · Accepted Answer

### The Geometry of Floating Points

Imagine you are standing in a vast, three-dimensional space. You have a straight line $L$ stretching out into infinity, and hovering above it, like a star in the night sky, is point $P(4, 2, 7)$.

We are tasked with finding two points, $Q$ and $R$, that lie on this line such that the distance from $P$ to both is exactly $\sqrt{26}$. This creates an isosceles triangle, $	riangle PQR$, suspended in space.

### The Strategy: The Foot of the Perpendicular

We could try to solve for $Q$ and $R$ by brute force, but that is a trap. Instead, let's use the most elegant tool in our arsenal: the foot of the perpendicular.

Imagine dropping a plumb line from $P$ straight down to the line $L$. Let the point where it touches the line be $M$. This point $M$ is the anchor of our entire solution.

By finding $M$, we split our isosceles triangle into two perfect, congruent right-angled triangles, $	riangle PQM$ and $	riangle PRM$. This simplifies our life immensely.

### Parametric Coordinates: The Key to the Line

To find $M$, we must first understand the line $L$. We are given the equation:

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

We introduce a parameter $\lambda$ to represent any point on this line. By setting the equation equal to $\lambda$, we can express any point on the line as $(2\lambda - 1, 3\lambda - 2, 2\lambda + 1)$.

This is our general coordinate for $M$. It is a variable point, waiting for us to find the specific $\lambda$ that makes it the foot of the perpendicular.

### The Dot Product Magic

Now, we need to ensure that the line segment $PM$ is perpendicular to the line $L$. We calculate the direction vector $\vec{PM}$ by subtracting the coordinates of $P$ from our parametric $M$:

$$\vec{PM} = (2\lambda - 5, 3\lambda - 4, 2\lambda - 6)$$

The direction vector of the line $L$ is simply the denominator of its equation: $\vec{b} = (2, 3, 2)$. For $PM$ to be perpendicular to $L$, their dot product must be zero: $\vec{PM} \cdot \vec{b} = 0$.

This gives us the equation:

$$2(2\lambda - 5) + 3(3\lambda - 4) + 2(2\lambda - 6) = 0$$

Expanding this, we get $4\lambda - 10 + 9\lambda - 12 + 4\lambda - 12 = 0$, which simplifies to $17\lambda - 34 = 0$. Solving this, we find the magic value: $\lambda = 2$.

### Calculating the Dimensions

With $\lambda = 2$, we find the coordinates of $M$ by substituting back: $M = (2(2) - 1, 3(2) - 2, 2(2) + 1) = (3, 4, 5)$.

Now, the height of our triangle, the length $PM$, is just the distance between $P(4, 2, 7)$ and $M(3, 4, 5)$. Using the distance formula:

$$PM = \sqrt{(3-4)^2 + (4-2)^2 + (5-7)^2} = \sqrt{1 + 4 + 4} = 3$$

We know the hypotenuse $PQ = \sqrt{26}$ and the height $PM = 3$. Applying the Pythagorean theorem to $	riangle PQM$, we get $QM^2 + PM^2 = PQ^2$, which means $QM^2 + 9 = 26$.

Thus, $QM^2 = 17$, or $QM = \sqrt{17}$. Since $M$ bisects the base $QR$, the total length of the base is $QR = 2\sqrt{17}$.

### The Final Victory

The area of $	riangle PQR$ is:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2\sqrt{17}) 	imes 3 = 3\sqrt{17}$$

The problem asks for the square of the area, so we calculate:

$$(3\sqrt{17})^2 = 9 	imes 17 = 153$$

We have arrived at the solution, not by brute force, but by understanding the geometric soul of the problem. The final answer is **153**.

Animated Solution for Mathematics - Three Dimensional Geometry: Let $Q$ and $R$ be two points on the line $\frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z - 1}{2}$ at a distance $26$ from the point $P (4, 2, 7)$ . Then the square of the area of the triangle $P QR$ is ______.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Strategy: Foot of the Perpendicular

Parametric Coordinates of $M$

Direction Ratios of $P M$

Perpendicularity Condition

Solving for $λ$

Coordinates of $M$

Calculating Altitude $P M$

Finding Base Segment $QM$

Calculating $QM$ and $QR$

Final Area Calculation

The Sigma Insight: Equation of a Line in Space

The Geometry of Floating Points

The Strategy

Parametric Coordinates

The Dot Product Magic

Calculating the Dimensions

The Final Victory

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