Question

The plane $x+2y+z=6$ cuts $X,YandZ$ axes at $A,BandC$ if $x$ coordinate of circumcentre of $△ABC$ is $\alpha$, then $2\alpha$ is equal to ______

Answer 1

Given eq of plane

$x+2y+z=6$

$\frac{x}{6}+\frac{y}{3}+\frac{z}{6}=1$

Comparing above eq with eq of plane in intercept form $\frac{x}{x_1}+\frac{y}{y_1}+\frac{z}{z_1}=1$

$x_1=6$

$y_1=3,z_1=6$

So coordinates of Points A,B,C

$A(6,0,0),B(0,3,0),C(0,0,6)$

let circumcentre $D(x,y,z)$

The distance of circumcentre from A,B,C should be same

$AD=\sqrt{(x-6{)}^2+y^2+z^2}$

$BD=\sqrt{x^2+(y-3{)}^2+z^2}$

$CD=\sqrt{x^2+y^2+(z-6{)}^2}$

$AD=BD=CD$

taking

$AD=BD$

$\sqrt{(x-6{)}^2+y^2+z^2}=\sqrt{x^2+(y-3{)}^2+z^2}$

on squaring bith sides

$(x-6{)}^2+y^2+z^2=x^2+(y-3{)}^2+z^2$

$x^2+y^2+z^2+36-12x=x^2+y^2+z^2+9-6y$

$36-12x=9-6y$

$12x-6y=27$

$4x-2y=9$-----(1)

from eq (1) x,y can be

$x=\frac{3}{2}$

$y=-\frac{3}{2}$

$\alpha =\frac{3}{2}$

$2\alpha =2\times \frac{3}{2}$

$2\alpha =3$