Question

The equation of the triangular plane of tetrahedron that contains the given vertices $A(6,-5,-1),B(-4,1,3)$ & $C(2,-4,18)$ is

• A. $55x+87y+7z=112$
• B. $55x+87y+7z+112=0$
• C. $55x-87y-7z=112$
• D. None of these

Answer 1

Answer: (B)

$55x+87y+7z+112=0$

Given vertices of tetrahedron are $(2,-4,18),(6,-5,-1)$ and $(-4,1,3)$
Let DC's of the normal to the plane are $(a,b,c)$.

So equation of plane through $(2,-4,18)$ is given by,
$a(x-2)+b(y+4)+c(z-18)=0$.........$\left(1\right)$
Now (1) passes through $A(6,-5,-1)$ and $B(-4,1,3)$
$\therefore$ $4a-b-19c=0$ & $6a-5b+15c=0$
$\therefore$ $\frac{a}{110}=\frac{b}{174}=\frac{c}{14}$
$\therefore$ By $\left(1\right)$ we get the equation of plane
$55x+87y+7z+112=0$