Question

A plane $P$ is parallel to two lines whose direction ratios are $-2,1,-3$, and $-1,2,-2$ and it contains the point $(2,2,-2)$. Let $P$ intersect the co-ordinate axes at the points $A$, $B,C$ making the intercepts $\alpha ,\beta ,\gamma$. If $V$ is the volume of the tetrahedron $OABC$, where $O$ is the origin and $p=\alpha +\beta +\gamma$, then the ordered pair $(V,p)$ is equal to

• A. $(48,-13)$
• B. $(24,-13)$
• C. $(48,11)$
• D. $(24,-5)$

Answer 1

Answer: (B)

Normal of plane $P$ :
$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& 1& -3\\ -1& 2& -2\end{array}\right|=4\hat{i}-\hat{j}-3\hat{k}$
Equation of plane $P$ which passes through (2, $2,-2)$ is $4x-y-3z-12=0$
Now, A $(3,0,0)$, B $(0,-12,0)$, C $(0,0,-4)$
$\Rightarrow \alpha =3,\beta =-12,\gamma =-4$
$\Rightarrow p=\alpha +\beta +\gamma =-13$
Now, volume of tetrahedron $OABC$
$V=\left|\frac{1}{6}\overrightarrow{OA}\cdot (\overrightarrow{OB}\times \overrightarrow{OC})\right|=24$
$(V,p)=(24,-13)$\