Question

Let $a_i,i=1,2,3,\dots ,n$ denote the integers in the domain of function $f\left(x\right)=\sqrt{{\log }_{1/2}\left(\frac{4x-25}{x-21}\right)}$ where $a_i<a_{i+1}\forall i\in N.$ A line $L:\frac{2x-a_1}{4}=\frac{y+a_1}{a_2}=\frac{z-a_3}{a_5}$ meets the $xy,yz$ and $zx$ planes at $A,B$ and $C$ respectively. If volume of the tetrahedron $OABD$ is $V$ cubic units where $O$ is origin and $D$ is the image of $C$ with respect to $x-$axis, then the value of $90V$ is

Answer 1

For domain of $f\left(x\right),$ we must have
${\log }_{1/2}\left(\frac{4x-25}{x-21}\right)\ge 0\Rightarrow \frac{4x-25}{x-21}\le 1\Rightarrow \frac{3x-4}{x-21}\le 0\Rightarrow x\in [\frac{4}{3},21)\dots \left(1\right)$

And $\frac{4x-25}{x-21}>0$
$\Rightarrow x\in (-\infty ,\frac{25}{4})\cup (21,\infty )\dots \left(2\right)$
From $\left(1\right)\cap \left(2\right),$ $x\in [\frac{4}{3},\frac{25}{4}]$
$\therefore$ Integers in the domain are $2,3,4,5,6$

$\therefore L:\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-4}{6}=r$ (let)
In $xy-$plane, $z=0$
$\Rightarrow 6r+4=0$
$\Rightarrow r=\frac{-2}{3}$
$\therefore A(2r+1,3r-2,0)\equiv (-\frac{1}{3},-4,0)$

In $yz-$plane, $x=0$
$\Rightarrow 2r+1=0$
$\Rightarrow r=\frac{-1}{2}$
$\therefore B(0,3r-2,6r+4)\equiv (0,\frac{-7}{2},1)$

In $zx-$plane, $y=0$
$\Rightarrow 3r-2=0\Rightarrow r=\frac{2}{3}$
$\Rightarrow C(2r+1,0,6r+4)\equiv (\frac{7}{3},0,8)$

Given that image of $C$ w.r.t. $x-$axis is $D$
$\Rightarrow D\equiv (\frac{7}{3},0,-8)$


$\therefore$ Volume of the tetrahedron
$OABD$ is $V=\frac{1}{6}\left|\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]\right|$
$V=\frac{1}{6}\left|\left|\begin{array}{ccc}-1/3& -4& 0\\ 0& -7/2& 1\\ 7/3& 0& -8\end{array}\right|\right|=\frac{28}{9}$
Hence, $90V=280$