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

Let $P=(7,5,2)$ Eq. of line passing through $P$ and parallel to given line is $$\frac{x-7}{3}=\frac{y-5}{6}=\frac{z-2}{2}=r\text{ (say) }$$ Hence coordinates of any point on this line are $(3r+7,6r+5,2r+2)\equiv Q($ say $)$ We have $3x+4y+z-8=0$ $\therefore 3(3r+7)+4(6r+5)+(2r+2)-8=0$ $\therefore 9r+24r+2r+21+20+2-8=0\Rightarrow 35r=-35\Rightarrow r=-1$ $\therefore Q\equiv (-3+7,-6+5,-2+2)$ i.e. $(4,-1,0)$ Distance between $\mathrm{PQ}\equiv \sqrt{(7-4{)}^2+(5+1{)}^2+(2-0{)}^2}=\sqrt{9+36+4}=7$