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

The point $\mathrm{A}(6,7,7)$ is on the line. Let the perpendicular from $\mathrm{P}$ meet the line in $\mathrm{L}$. Then ${\mathrm{AP}}^2=(6-1{)}^2+(7-2{)}^2+(7-3{)}^2=66$ Also $\mathrm{AL}=$ projection of AP on line
$\begin{array}{rl}& \left(\text{actual d.c.’s }\frac{3}{\sqrt{17}},\frac{2}{\sqrt{17}},\frac{-2}{\sqrt{17}}\right)\\ \Rightarrow & (6-1)\cdot \frac{3}{\sqrt{17}}+(7-2)\cdot \frac{2}{\sqrt{17}}+(7-3)\frac{-2}{\sqrt{17}}\\ & =\sqrt{17}\end{array}$ $\therefore \perp$ distance $\mathrm{d}$ of $\mathrm{P}$ from the line is given by ${\mathrm{d}}^2={\mathrm{AP}}^2-{\mathrm{AL}}^2=66-17=49\text{ so that }\mathrm{d}=7$