Mathematics - Vector Question with Solution | TestHub

$\mathrm{R}(-1+2\mathrm{r},3-2\mathrm{r},-\mathrm{r})$

$dr\text{'}s of PR are (2-2 r,-1+2 r,-3+r)$

Then,  $2(2-2 r)+2(1-2 r)+1(3-r)=0$

$9-9r=0 \Rightarrow r=1$

$R(1,1,-1)$

then  $a+1=2 ,b+2=2 , c-3=-2$ i.e., $\mathrm{a}=1,\mathrm{b}=0, \mathrm{c}=1$

$\therefore a+b+c=2$