Let P(1,6,3) be the given point and L be the foot of perpendicular from P to the given line. The coordinates of a general point on the given line are
x−0
1
=
y−1
2
=
z−2
3
=λ (say) i.e. x=λy=2λ+1,z=3λ+2 If the coordinates of L are (λ,2λ+1,3λ+2), then the direction ratios of PL are (λ−1,2λ−5,3λ−1). Since, the direction ratios of given line which is perpendicular to PL, are 1,2 and 3 . Therefore, (λ−1)1+(2λ−5)2+(3λ−1)3=0, which gives λ=1 Hence, coordinates of L are (1,3,5). Let Q(x1,y1,z1) be the image of P(1,6,3) in the given line. Then, L is the mid-point of PQ. Therefore,
x1+1
2
=1,
y1+6
2
=3,
z1+3
2
=5 ⇒x1=1,y=0,z=7 Hence, the image of (1,6,3) in the given line is (1,0,7).