Let the plane x+3y-2z+6=0 meet the coordinate | vedclass

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(D) The plane equation is $x+3y-2z+6=0$. Setting two coordinates to zero,we find the intercepts: $A(-6, 0, 0)$,$B(0, -2, 0)$,$C(0, 0, 3)$. Let $H(\alp

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$288$

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(D) The plane equation is $x+3y-2z+6=0$. Setting two coordinates to zero,we find the intercepts: $A(-6, 0, 0)$,$B(0, -2, 0)$,$C(0, 0, 3)$. Let $H(\alpha, \beta, \frac{6}{7})$ be the orthocentre. Since $H$ lies on the plane $ABC$,$\alpha + 3\beta - 2(\frac{6}{7}) + 6 = 0 \implies \alpha + 3\beta = -6 + \frac{12}{7} = -\frac{30}{7}$. Also,$\overrightarrow{AH} \cdot \overrightarrow{BC} = 0$. $\overrightarrow{AH} = (\alpha+6, \beta, \frac{6}{7}-0) = (\alpha+6, \beta, \frac{6}{7})$. $\overrightarrow{BC} = (0, 2, 3)$. $(\alpha+6)(0) + \beta(2) + \frac{6}{7}(3) = 0 \implies 2\beta + \frac{18}{7} = 0 \implies \beta = -\frac{9}{7}$. Substituting $\beta$ into the plane equation: $\alpha + 3(-\frac{9}{7}) = -\frac{30}{7} \implies \alpha - \frac{27}{7} = -\frac{30}{7} \implies \alpha = -\frac{3}{7}$. Now,$98(\alpha+\beta)^2 = 98(-\frac{3}{7} - \frac{9}{7})^2 = 98(-\frac{12}{7})^2 = 98 	imes \frac{144}{49} = 2 	imes 144 = 288$.

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

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