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:

• A. 26
• B. 36
• C. 18
• D.

Answer 1

The Correct Option is C

We are given the line equation in symmetric form:

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

Let the parameter be $t$. Then, we can parametrize the coordinates of any point on the line as:

$x = 8t - 3$, $y = 2t + 4$, $z = 2t - 1$

Thus, the points P and Q on the line can be written as:

$P(8t_1 - 3, 2t_1 + 4, 2t_1 - 1)$, $Q(8t_2 - 3, 2t_2 + 4, 2t_2 - 1)$

Step 1: Distance from P and Q to R(1, 2, 3)

We are also told that both P and Q are at a distance of 6 units from the point R(1, 2, 3). The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

For the point P, the distance to R is:

$\sqrt{(8t_1 - 3 - 1)^2 + (2t_1 + 4 - 2)^2 + (2t_1 - 1 - 3)^2} = 6$

This simplifies to:

$\sqrt{(8t_1 - 4)^2 + (2t_1 + 2)^2 + (2t_1 - 4)^2} = 6$

Squaring both sides:

$(8t_1 - 4)^2 + (2t_1 + 2)^2 + (2t_1 - 4)^2 = 36$

Expanding each term:

$(64t_1^2 - 64t_1 + 16) + (4t_1^2 + 8t_1 + 4) + (4t_1^2 - 16t_1 + 16) = 36$

Simplifying:

$72t_1^2 - 72t_1 + 36 = 36$

$72t_1^2 - 72t_1 = 0$

$72t_1(t_1 - 1) = 0$

Thus, $t_1 = 0$ or $t_1 = 1$.

Similarly, for Q, we get the same equation, leading to the same values for $t_2$: $t_2 = 0$ or $t_2 = 1$.

Step 2: Coordinates of Points P and Q

For $t_1 = 0$, the coordinates of P are: P(-3, 4, -1)

For $t_1 = 1$, the coordinates of P are: P(5, 6, 1)

Similarly, for $t_2 = 0$, the coordinates of Q are: Q(-3, 4, -1)

For $t_2 = 1$, the coordinates of Q are: Q(5, 6, 1)

Step 3: Centroid of Triangle PQR

The centroid of a triangle with vertices at $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ is given by:

$\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right)$

For the case $t_1 = 0$ and $t_2 = 1$, the coordinates of the centroid are:

$\left( \frac{-3 + 5 + 1}{3}, \frac{4 + 6 + 2}{3}, \frac{-1 + 1 + 3}{3} \right) = \left( 1, 4, 1 \right)$

Step 4: Calculate $\alpha^2 + \beta^2 + \gamma^2$

The centroid is (1, 4, 1), so:

$\alpha^2 + \beta^2 + \gamma^2 = 1^2 + 4^2 + 1^2 = 18$

Thus, the value of $\alpha^2 + \beta^2 + \gamma^2$ is $\boxed{18}$