MCQ
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
    $(48,-13)$
  • $(24,-13)$
  • C
    $(48,11)$
  • D
    $(24,-5)$

Answer

Correct option: B.
$(24,-13)$
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 $4 x-y-3 z-12=0$

Now, A $(3,0,0)$, B $(0,-12,0)$, C $(0,0,-4)$ $\Rightarrow \alpha=3, \beta=-12, \gamma=-4$

$p =\alpha+\beta+\gamma=-13$

Now, volume of tetrahedron $OABC$

$V =\left|\frac{1}{6} \overline{ OA } \cdot(\overline{ OB } \times \overrightarrow{ OC })\right|=24$

$( V , p )=(24,-13)$

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