The point A(6,7,7) is on the line, let the perpendicular from P meet the line in L. then_{$A{P}^2={\left(6-1\right)}^2+{\left(7-2\right)}^2+{\left(7-3\right)}^2=66$}

Also AL=projection of AP on line (actual d.c’s _{$\frac{3}{\sqrt{17}},\frac{2}{\sqrt{17}},\frac{-2}{\sqrt{17}}$} )

_{$\Rightarrow \left(6-1\right).\frac{3}{\sqrt{17}}+\left(7-2\right).\frac{2}{\sqrt{17}}+\left(7-3\right)\frac{-2}{\sqrt{17}}=\sqrt{17}$}

_{$\therefore \perp$} distance d of P from the line is given by _{$d^2=A{P}^2-A{L}^2=66-17=49$} so that d = 7