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

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 :

Accepted Answer

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

can be taken as $(8 \lambda-3,2 \lambda+4,2 \lambda-1)$

If at a distance of 6 units from $R(1,2,3)$

$$
\begin{aligned}
& \Rightarrow(8 \lambda-3-1)^2+(2 \lambda+4-2)^2+(2 \lambda-1-3)^2=36 \\
& \left.\Rightarrow \lambda^2-\lambda=0 \text { \{on simplification }\right\} \\
& \Rightarrow \lambda=0, \lambda=1
\end{aligned}
$$

Here $P \& Q$ are $(-3,4,-1)$ and $(5,6,1)$ Centroid of $\triangle P Q R$

$$
\begin{aligned}
& (\alpha, \beta, \gamma) \equiv\left(\frac{5-3+1}{3}, \frac{6+4+2}{3}, \frac{1-1+3}{3}\right) \\
& \Rightarrow \alpha=1, \beta=4, \gamma=1 \\
& \Rightarrow \alpha^2+\beta^2+\gamma^2=18
\end{aligned}
$$

Solutions · 1