The image of the point ( 1, 6, 3 ) in the line (x)/(1) = (y -...

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

$\dfrac{x - 0}{1} = \dfrac{y - 1}{2} = \dfrac{z - 2}{3} = \lambda$ & (say)

i.e. $x = \lambda$

$y = 2\lambda + 1, z = 3\lambda + 2$

If the coordinates of $L$ are $( \lambda, 2\lambda + 1, 3\lambda + 2 )$, then the direction ratios of $PL$ are $( \lambda - 1, 2\lambda - 5, 3\lambda - 1 )$.

Since, the direction ratios of given line which is perpendicular to $PL$, are $1, 2$ and $3$. Therefore, $( \lambda - 1 ) 1 + ( 2\lambda - 5 ) 2 + ( 3\lambda - 1 )^{3} = 0$,which gives $\lambda = 1$.

Hence, coordinates of $L$ are $( 1, 3, 5 )$.

Let $Q \left({x}_{1}, {y}_{1}, {z}_{1}\right) be the image of$P ( 1, 6, 3 )$in the given line. Then,$L$is the mid-point of$PQ$.</p><p>Therefore, \dfrac{{x}_{1} + 1}{2} =1, \dfrac{{y}_{1} + 6}{2} =3, \dfrac{{z}_{1} + 3}{2} =5$

$⇒ {x}_{1} = 1, y = 0, z = 7$

Hence, the image of $( 1, 6, 3 )$ in the given line is $( 1, 0, 7 )$.