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

Let $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}=\mathrm{k}$ (say) Let $\mathrm{P}$ be the any point on the above line. $\therefore \mathrm{P}=(2\mathrm{k}+1,3\mathrm{k}-1,4\mathrm{k}+2)$ The point $\mathrm{P}$ lies on the plane $\begin{array}{ll}\therefore & 2\mathrm{k}+1+2(3\mathrm{k}-1)+3(4\mathrm{k}+2)=15\\ \therefore & 2\mathrm{k}+1+6\mathrm{k}-2+12\mathrm{k}+6=15\\ \therefore & 20\mathrm{k}=10\\ \therefore & \mathrm{k}=\frac{1}{2}\\ \therefore & \mathrm{P}=\left(2,\frac{1}{2},4\right)\end{array}$ Distance of $\mathrm{P}$ from origin is $\sqrt{2^2+{\left(\frac{1}{2}\right)}^2+4^2}=\sqrt{\frac{81}{4}}=\frac{9}{2}$