Question

Let $A(1,1,1),B(2,3,5)$ and $C(-1,0,2)$ be three points, then equation of a plane parallel to the plane $ABC$ and at the distance $2$ is

• A. $2x-3y+z-2\sqrt{14}=0$
• B. $2x-3y+z-\sqrt{14}=0$
• C. $2x-3y+z+2=0$
• D. $2x-3y+z-2=0$

Answer 1

The correct option is A $2x-3y+z-2\sqrt{14}=0$

$\overrightarrow{AB}=i+2j+4k$
$\overrightarrow{BC}=3i+3j+3k$

Hence, $\overrightarrow{AB}\times \overrightarrow{BC}=3(2i-3j+k)$

Now the unit normal of the plane of $ABC$ will be

$=\frac{2i-3j+k}{\sqrt{14}}$

The required plane is parallel to the plane $ABC$.

Hence, its unit normal will be parallel to the normal of $ABC$.

Therefore, the equation of the required plane is

$r.\left(\frac{2i-3j+k}{\sqrt{14}}\right)=d$

$2x-3y+z=d\sqrt{14}$

Now $d$ is $2$.

Hence, the equation is $2x-3y+z=2\sqrt{14}$.