Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Given,

Equation of plane is

$\overrightarrow{r}=-\left(t+p\right)\hat{i}+t\hat{j}+\left(1+p\right)\hat{k}$

On rearranging we get,

$\overrightarrow{r}=\hat{k}+t\left(-\hat{i}+\hat{j}\right)+p\left(-\hat{i}+\hat{k}\right)$

So, equation of plane in standard form is given by

$\left[\begin{array}{lll}\overrightarrow{r}-\hat{k}& -\hat{i}+\hat{j}& -\hat{i}+\hat{k}\end{array}\right]=0$

$\Rightarrow \left|\begin{array}{ccc}x& y& z-1\\ -1& 1& 0\\ -1& 0& 1\end{array}\right|=0$

$\Rightarrow x+y+z=1 \cdots \left(1\right)$

Now given coordinate of $Q=\left(10,15,20\right)$

And coordinates of $S=\left(\alpha ,\beta ,\gamma \right)$

Now using the image formula of point and plane we get,

$\frac{\alpha -10}{1}=\frac{\beta -15}{1}=\frac{\gamma -20}{1}=\frac{-2\left(10+15+20-1\right)}{3}$

$\Rightarrow \alpha -10=\beta -15=\gamma -20=-\frac{88}{3}$

$\Rightarrow \alpha =-\frac{58}{3},\beta =-\frac{43}{3},\gamma =-\frac{28}{3}$

Now solving all options we get,

$3\left(\alpha +\beta \right)=-101$

$3\left(\beta +\gamma \right)=-71$

$3\left(\gamma +\alpha \right)=-86$ 

$3\left(\alpha +\beta +\gamma \right)=-129$