Question

Find the coordinates of those points on the line $\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-3}{6}$ which is at a distance of $3$ units from the point $(1,-2,3)$

Answer 1

$\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-3}{6}$

Let the point be

$=(2k-1,3k-2,6k+3)$

Distance from $(1,-2,3)$

$=\sqrt{(2k-1-1{)}^2+(3k-2+2{)}^2+(6k+3-3{)}^2}$

$=\sqrt{(2k-2^2+9k^2+36k^2}=3$

$\therefore 4k^2+4-8k+45k^2=9$

$\therefore 49k^2-8k-5=0$

$\therefore k=\frac{8\pm \sqrt{64+20\times 49}}{98}$

$\therefore k=\frac{8\pm \sqrt{65+980}}{98}$

$\therefore k=\frac{8\pm 32}{98}=\frac{40}{98}=\frac{20}{49}$

$\therefore$ Point are $(\frac{-9}{49},\frac{-38}{49},\frac{267}{49})$