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The equation of the plane which makes double the intercepts of the plane $3x+6y-z+12=0$ is ?

$(a)\:6x+12y-2z=1\:\:\qquad\:(b)\:3x+6y-z+6=0\:\:\qquad\:(c)\:3x+6y-z=1\:\:\qquad\:(d)\:3x+6y-z+24=0$

1 Answer

The given plane is $3x+6y-z+12=0$
Converting the eqn. into intercept form $\large\frac{x}{-4}+\frac{y}{-2}+\frac{z}{12}=1$
The intercepts of this plane are $ (-4,-2,12)$
The intercepts of the required plane are $(-8,-4,24)$
$\therefore\:$ The eqn. of the required plane is $\large\frac{x}{-8}+\frac{y}{-4}+\frac{z}{24}=1$
$i.e.,\:\:3x+6y-z+24=0$
answered Jan 7, 2014 by rvidyagovindarajan_1
 

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