JEE MAIN - Mathematics (2024 - 1st February Evening Shift - No. 15)
Let $\mathrm{P}$ and $\mathrm{Q}$ 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 $\mathrm{R}(1,2,3)$. If the centroid of the triangle PQR is $(\alpha, \beta, \gamma)$, then $\alpha^2+\beta^2+\gamma^2$ is :
18
24
26
36
Explanation
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 $\triangle P Q R$
$$ \begin{aligned} & (\alpha, \beta, \gamma) \equiv\left(\frac{5-3+1}{3}, \frac{6+4+2}{3}, \frac{1-1+3}{3}\right) \\\\ & \Rightarrow \alpha=1, \beta=4, \gamma=1 \\\\ & \Rightarrow \alpha^2+\beta^2+\gamma^2=18 \end{aligned} $$
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 $\triangle P Q R$
$$ \begin{aligned} & (\alpha, \beta, \gamma) \equiv\left(\frac{5-3+1}{3}, \frac{6+4+2}{3}, \frac{1-1+3}{3}\right) \\\\ & \Rightarrow \alpha=1, \beta=4, \gamma=1 \\\\ & \Rightarrow \alpha^2+\beta^2+\gamma^2=18 \end{aligned} $$
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