Let A(1, 6, 3) be the given point and set L be the foot of perpendicular from P to given line.
The co-ordinates of a general point on the line are
\(\frac{x-0}{1} = \frac{y-1}{2}= \frac{z-2}{3}= \lambda\)
i.e. \(x= \lambda, \ y = 2\lambda +1, z = 3\lambda +2\)
If the co-ordinates of L are
\((\lambda,\ 2\lambda+1, 3\lambda+2), \) then direction ratio of PL are \(<\lambda -1 ,2\lambda - 5, 3\lambda - 1>\)
But the direction ratio of given line which is perpendicular to AL are \(<1,2,3>\)
\(\therefore (\lambda -1)+ 2(2\lambda - 5)+ 3(3\lambda-1) = 0\)
\(\Rightarrow x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) \(\lambda = 1\)
\(\therefore\) L (1, 3, 5)
\(\because\) L is the mid pointing AA'
\(\therefore \frac{x_1+1}{2} = 1 \) \(\Rightarrow x_1 = 1\)
\(\frac{y_1+6}{2} = 3 \) \(\Rightarrow y_1 = 0\)
\(\frac{z_1+ 3}{2} = 5\) \(\Rightarrow z_1 = 7\)
\(\therefore\)image of (1, 6, 3) in given line is (1,0,7)
Line joining A an A' will have direction ratio, < 0, 6, –4 >
\(\therefore\) equation of line will be
\(\frac{x-1}{0}= \frac{y-6}{6} = \frac{z-3}{-4}\)
