Question Number 56696 by Tawa1 last updated on 21/Mar/19
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{perpendicular}\:\mathrm{distance}\:\mathrm{from}\:\:\left(\mathrm{1},\:\mathrm{7},\:\mathrm{1}\right)\:\:\mathrm{to}\:\:\mathrm{3x}\:−\:\mathrm{2y}\:+\:\mathrm{2z}\:\:=\:\:\mathrm{6} \\ $$
Commented by mr W last updated on 21/Mar/19
$${d}=\frac{\mid\mathrm{3}×\mathrm{1}−\mathrm{2}×\mathrm{7}+\mathrm{2}×\mathrm{1}−\mathrm{6}\mid}{\sqrt{\mathrm{3}^{\mathrm{2}} +\left(−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}=\frac{\mathrm{15}}{\sqrt{\mathrm{17}}} \\ $$
Commented by Tawa1 last updated on 22/Mar/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$