BITSAT 2017 Slot 2 Solved Paper

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 $\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda$ (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 $P L$ are $(\lambda-1, 2\lambda-5, 3\lambda-1)$. Since, the direction ratios of given line which is perpendicular to $P L$, 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 $Q(x_{1}, y_{1}, z_{1})$ be the image of $P(1,6,3)$ in the given line. Then, $L$ is the mid-point of $P Q$. Therefore, $\frac{x_{1}+1}{2}=1, \frac{y_{1}+6}{2}=3, \frac{z_{1}+3}{2}=5$ $\Rightarrow x_{1}=1, y=0, z=7$ Hence, the image of $(1,6,3)$ in the given line is $(1,0,7)$.