Let the plane $x+3 y-2 z+6=0$ meet the coordinate axes at the points $A, B, C$. If the orthocenter o

Question

Let the plane $x+3 y-2 z+6=0$ meet the coordinate axes at the points $A, B, C$. If the orthocenter of the triangle $A B C$ is $\left(\alpha, \beta, \frac{6}{7}ight)$, then $98(\alpha+\beta)^{2}$ is equal to $\_\_\_\_$ .\

Professor PC Thomas Classes · Answer

(288): Plane $x+3 y-2 z+6=0$ meets the coordinate axes at the points $A, B, C$.\ For $x$-axis; $x+0-0+6=0 \Rightarrow x=-6$ i.e., $(-6,0,0)$\ For $y$-axis; $0+3 y-0+6=0 \Rightarrow y=-2$ i.e., ( $0,-2,0$ )\ For $z$-axis; $0+0-2 z+6=0 \Rightarrow z=3$ i.e., $(0,0,3)$\ $\overrightarrow{A B}=6 \hat{i}-2 \hat{j} ; \overrightarrow{B C}=2 \hat{j}+3 \hat{k}$;\ $\overrightarrow{A C}=6 \hat{i}+3 \hat{k}$\ Now, $\overrightarrow{A P} \cdot \overrightarrow{B C}=0$\ $	herefore \quad\left(\alpha+6, \beta, \frac{6}{7}ight) \cdot(0,2,3)=0$\ $\Rightarrow \quad 2 \beta+3 \cdot \frac{6}{7}=0 \Rightarrow \beta=\frac{-9}{7}$\ Similarly, $\overrightarrow{C P} \cdot \overrightarrow{A B}=0$\ $\left(\alpha, \beta, \frac{-15}{7}ight) \cdot(6,-2,0)=0 \Rightarrow 6 \alpha-2 \beta=0$\ $\Rightarrow \quad 6 \alpha-2\left(\frac{-9}{7}ight)=0 \Rightarrow 6 \alpha=\frac{-18}{7}$\ $\Rightarrow \quad \alpha=\frac{-3}{7}$\ $	herefore \quad 98(\alpha+\beta)^{2}=98\left(\frac{-3}{7}-\frac{9}{7}ight)^{2}$\ $=98\left(\frac{-12}{7}ight)^{2}=98 \cdot \frac{144}{49}=288$\