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Let the food of perpendicular from the point \mathrm{A}(4,3,1) on the plane \mathrm{P}: \mathrm{x}-\mathrm{y}+2 \mathrm{z}+3=0 be \mathrm{N}. If \mathrm{B}(5, \alpha, \beta)$, $\alpha, \beta \in \mathrm{Z} is a point on plane \mathrm{P} such that the area of the triangle A B N is 3 \sqrt{2}, then \alpha^2+\beta^2+\alpha \beta is equal to __________.

Option: 1

7


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

$$ \begin{aligned} & \mathrm{AN}=\sqrt{6} \\ & 5-\alpha+2 \beta+3=0 \\ & \Rightarrow \alpha=8+2 \beta \end{aligned}\\ $$ \mathrm{N}$ is given by $$

\begin{aligned} & \frac{x-4}{1}=\frac{y-3}{-1}=\frac{z-1}{2}=\frac{-(4-3+2+3)}{1+1+4} \\ & \mathrm{x}=3, \mathrm{y}=4, \mathrm{z}=-1 \\ & \mathrm{~N} \end{aligned}

\begin{aligned} & (3,4,-1) \\ & \mathrm{BN}=\sqrt{4+(\alpha-4)^2+(\beta+1)^2} \\ & =\sqrt{4+(2 \beta+4)^2+(\beta+1)^2} \\ & \text { Area of } \triangle \mathrm{ABN}=\frac{1}{2} \mathrm{AN} \times \mathrm{BN}=3 \sqrt{2} \end{aligned}

\begin{aligned} & \frac{1}{2} \times \sqrt{6} \times \mathrm{BN}=3 \sqrt{2} \\ & \mathrm{BN}=2 \sqrt{3} \\ & 4+(2 \beta+4)^2+(\beta+1)^2=12 \\ & (2 \beta+4)^2+(\beta+1)^2-8=0 \\ & 5 \beta^2+18 \beta+9=0 \\ & (5 \beta+3)(\beta+3)=0 \\ & \beta=-3 \\ & \alpha=2 \\ & \alpha^2+\beta^2+\alpha \beta=9+4-6=7 \end{aligned}

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rishi.raj

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