Let \[\mathrm{P}(\alpha, \beta, \gamma)\] be the image of the point \[\mathrm{Q}(3,-3,1)\] in the line \[\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}\] and \[\mathrm{R}\] be the point (2,5,−1). If the area of the triangle \[\mathrm{PQR}\] is \[\lambda\] and \[\lambda^2=14 \mathrm{~K}\], then \[\mathrm{K}\] is equal to :
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Official solution
\[\text { Line } L \Rightarrow \frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}=\mu\]
Point \[M(\mu, \mu+3,-\mu+1)\]
Direction vector of \[Q M=(\mu-3, \mu+6,-\mu)\]
Direction vector of line L=(1,1,−1)
Both direction vector are perpendicular so
\[1×(μ−3)+1×(μ+6)−1×(−μ)=0μ=−1\]
Point M(−1,2,2) midpoint of PQ
So point \[P(\alpha, \beta, \gamma)=(-5,7,3)\]
Area of \[\triangle P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{Q R}|\]
\[\begin{aligned}
& \overrightarrow{P Q}=(8,-10,-2) \\
& \overrightarrow{Q R}=(-1,8,-2) \\
& \overrightarrow{P Q} \times \overrightarrow{Q R}=\left|\[\begin{array}{|ccccccc|}\hat{i} & \hat{j} & \hat{k} \\ 8 & -10 & -2 \\ -1 & 8 & -2\end{array}\end{aligned}\]\right|=36 \hat{i}+18 \hat{j}+54 \hat{k}
\]
Area of \[\triangle P Q R=\lambda\]
\[\lambda=\frac{1}{2} \sqrt{(36)^2+(18)^2+(54)^2}\]
\[λ2=41(4536)λ2=1134λ2=14K (given) 14K=1134K=141134=81K=81\]