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 \[\mathrm{A}, \mathrm{B}, \mathrm{C}\] making the intercepts \[\alpha, \beta, \gamma\]. If \[\mathrm{V}\] is the volume of the tetrahedron \[\mathrm{OABC}\], where \[\mathrm{O}\] is the origin, and \[\mathrm{p}=\alpha+\beta+\gamma\], then the ordered pair \[(\mathrm{V}, \mathrm{p})\] is equal to :
Solutions · 1
Official solution
Let \[{\overrightarrow a _1} = ( - 2,1, - 3)\] and \[{\overrightarrow a _2} = ( - 1,2, - 2)\]
Vector normal to plane \[\overline n = {\overrightarrow a _1} \times {\overrightarrow a _2}\]
\[\overline n = (4, - 1, - 3)\]
Plane through (2,2,−2) and normal to \[\overline n \]
\[(x - 2,y - 2,z + 2)\,.\,(4, - 1, - 3) = 0\]
\[ \Rightarrow 4x - y - 3z = 12\]
\[ \Rightarrow {x \over 3} + {y \over { - 12}} + {z \over { - 4}} = 1\]
Intercepts \[\alpha\], \[\beta\], \[\gamma\] are 3,−12,−4
\[P = \alpha + \beta + \gamma = - 13\]
\[V = {1 \over 6} \times 3 \times 12 \times 4 = 24\]