Let P and Q be the points on the line (x+3)/8 = (y-4)/2 = (z+1)/2 which are at a distance of 6 units

Q: Why do we use $\lambda$ instead of just solving for $x, y, z$ directly?

Think of $\lambda$ as a 'coordinate locator'. A line in 3D space contains infinite points. By equating the line to $\lambda$, we collapse the 3D complexity into a single 1D parameter. This allows us to express any point on the line as a function of $\lambda$, turning a geometry problem into a simple algebraic one.

Q: Why square the distance? Isn't it just $PR = 6$?

You are absolutely right that $PR = 6$. However, the distance formula involves a square root: $\sqrt{(x_P - x_R)^2 + ...} = 6$. Squaring both sides gives $(x_P - x_R)^2 + ... = 36$. This removes the radical, making the algebra much cleaner and significantly reducing the chance of making a sign error during expansion.

Q: What if the quadratic equation for $\lambda$ had no real roots?

That is a fantastic conceptual question! If the quadratic had no real roots (i.e., a negative discriminant), it would mean the line never gets as close as 6 units to point $R$. It would imply the line passes entirely outside the sphere of radius 6 centered at $R$. But here, the math works out perfectly, confirming the line intersects that sphere at two distinct points.

Question

Let $P$ and $Q$ be the points on the line $\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}$ which are at a distance of 6 units from the point $R(1, 2, 3)$. If the centroid of the triangle $PQR$ is $(\alpha, \beta, \gamma)$, then $\alpha^2 + \beta^2 + \gamma^2$ is:

Options:

(A) 26

(B) 36

(C) 18

(D) 24

Sigma Percentile · Accepted Answer

### The Geometry of the Path

Imagine you are standing in a vast, three-dimensional room. You see a straight line stretching infinitely in both directions, defined by the equation:

$$
\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}
$$

Floating nearby is a point $R(1, 2, 3)$. We are tasked with finding two specific points, $P$ and $Q$, that lie on this line, both exactly 6 units away from $R$.

### The Parametric Bridge

To capture a point on the line, we use a parameter, $\lambda$. By setting the line equation equal to $\lambda$, we unlock the ability to describe any point on the line as a function of this single variable.

This gives us the coordinates of a general point $P$ as:

$$
x = 8\lambda - 3, \quad y = 2\lambda + 4, \quad z = 2\lambda - 1
$$

This is our bridge between the abstract line and concrete algebra.

### The Distance Dance

We know the distance $PR$ is 6. To simplify our calculations, we use the squared distance formula $PR^2 = 36$.

Substituting our parametric coordinates for $P$ and the fixed coordinates for $R(1, 2, 3)$, we get the equation:

$$
(8\lambda - 3 - 1)^2 + (2\lambda + 4 - 2)^2 + (2\lambda - 1 - 3)^2 = 36
$$

Simplifying the terms inside the brackets, we arrive at:

$$
(8\lambda - 4)^2 + (2\lambda + 2)^2 + (2\lambda - 4)^2 = 36
$$

### The Algebraic Unfolding

Expanding these squares gives us:

$$
(64\lambda^2 - 64\lambda + 16) + (4\lambda^2 + 8\lambda + 4) + (4\lambda^2 - 16\lambda + 16) = 36
$$

When we group the like terms, the equation collapses into:

$$
72\lambda^2 - 72\lambda + 36 = 36
$$

The constant 36 cancels out perfectly from both sides, leaving us with $72\lambda^2 - 72\lambda = 0$. Factoring this, we get $72\lambda(\lambda - 1) = 0$, which yields two elegant solutions: $\lambda = 0$ and $\lambda = 1$.

### The Final Centroid

With $\lambda = 0$, we find $P = (-3, 4, -1)$. With $\lambda = 1$, we find $Q = (5, 6, 1)$.

The centroid $G(\alpha, \beta, \gamma)$ of triangle $PQR$ is the average of the coordinates of its vertices:

$$
\alpha = \frac{-3+5+1}{3} = 1, \quad \beta = \frac{4+6+2}{3} = 4, \quad \gamma = \frac{-1+1+3}{3} = 1
$$

Finally, we calculate the requested value:

$$
\alpha^2 + \beta^2 + \gamma^2 = 1^2 + 4^2 + 1^2 = 1 + 16 + 1 = \mathbf{18}
$$

Select Answer:

Visualized Solution

Visualizing the Problem

Parametric Form of the Line

Coordinates of a General Point

The Distance Constraint

Setting up the Equation

Simplifying the Brackets

Expanding the Quadratics

Combining Like Terms

Solving for Lambda

Finding Points P and Q

The Centroid Formula

Calculating Centroid Coordinates

Final Computation

The Sigma Insight: Equation of a Line in Space

The Geometry of the Path

The Parametric Bridge

The Distance Dance

The Algebraic Unfolding

The Final Centroid

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