Explanation:
D Given,
\(\mathrm{A}=(1,2,3)\)
line, \(\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}=k\)
(say)
So, any point on this line is
\((3 \mathrm{k}+6,2 \mathrm{k}+7,-2 \mathrm{k}+7)\)
DR of a line which is perpendicular to this line will be
\((3 \mathrm{k}+6-1,2 \mathrm{k}+7-2,-2 \mathrm{k}+7-3)\)
\(\Rightarrow (3 \mathrm{k}+5,2 \mathrm{k}+5,-2 \mathrm{k}+4)\)
\(\Rightarrow \text { D.R. of given line are }(3,2,-2) \ldots . \text { (iii) }\)
\(\Rightarrow\) D.R. of given line are \((3,2,-2) \ldots\) (iii)
From equation(ii) and (iii) are perpendicular to each other,
\(\Rightarrow 3(3 \mathrm{k}+5)+2(2 \mathrm{k}+5)-2(-2 \mathrm{k}+4) =0\)
\(9 \mathrm{k}+15+4 \mathrm{k}+10+4 \mathrm{k}-8 =0\)
\(17 \mathrm{k} =-17\)
\( \mathrm{k} =-1\)
substituting \(\mathrm{k}=-1\) in equation (i)-
Point is \((3,5,9)\)
Now, distance between \((3,5,9)\) and \((1,2,3)\) will be
\(d =\sqrt{4+9+36}\)
\(d =\sqrt{49}\)
\(\Rightarrow \quad =7\)