QJEEM24Ma373e: JEE Main 2024 Mathematics - Class 12 JEEM24Ma373e Solution: Let and be the points on the line which are at a distance of 6…

Question

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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 :

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 λ - 3 - 1)^{2} + (2 λ + 4 - 2)^{2} + (2 λ - 1 - 3)^{2} = 36 \\ \Rightarrow λ^{2} - λ = 0 {on simplification} \\ \Rightarrow λ = 0, λ = 1 \end{aligned}

Here

P \&amp; Q

are

(-3,4,-1)

and

(5,6,1)

Centroid of

\triangle P Q R

\begin{aligned} (α, β, γ) \equiv (\frac{5 - 3 + 1}{3}, \frac{6 + 4 + 2}{3}, \frac{1 - 1 + 3}{3}) \\ \Rightarrow α = 1, β = 4, γ = 1 \\ \Rightarrow α^{2} + β^{2} + γ^{2} = 18 \end{aligned}

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