Let the image of the point ( 1, 0, 7 ) in the line (x)/(1) = (y -...

Given: 

${L}_{1} ≡ \dfrac{x}{1} = \dfrac{y - 1}{2} = \dfrac{z - 2}{3} = \lambda$

$⇒ M ( \lambda, 1 + 2\lambda, 2 + 3\lambda )$

⇒ \overset{\to}{PM} = \left(\lambda - 1\right)\overset{^}{i} + \left(1 + 2\lambda\right)\overset{^}{j} + \left(3\lambda - 5\right)\overset{^}{k}

So, $\overset{\to}{PM}$ is perpendicular to line ${L}_{1}$

$⇒ \overset{\to}{PM} \cdot \overset{\to}{b} = 0$

⇒ \overset{\to}{PM} . \overset{\to}{b} = \left[\left(\lambda - 1\right)\overset{^}{i} + \left(1 + 2\lambda\right)\overset{^}{j} + \left(3\lambda - 5\right)\overset{^}{k}\right] . \left(\overset{^}{i} + 2\overset{^}{j} + 3\overset{^}{k}\right)

$⇒ \lambda - 1 + 4\lambda + 2 + 9\lambda - 15 = 0$

$⇒ 14\lambda = 14$

$⇒ \lambda = 1$

$⇒ M = ( 1, 3, 5 )$

Now, $M$ is midpoint of $\overset{\to}{P}$ and $\overset{\to}{Q}$

$⇒ \left(1, 3, 5\right) ≡ \left(\dfrac{1 + \alpha}{2}, \dfrac{0 + \beta}{2}, \dfrac{7 + \gamma}{2}\right)$

$⇒ \dfrac{1 + \alpha}{2} = 1, \dfrac{\beta}{2} = 3, \dfrac{7 + \gamma}{2} = 5$

$⇒ \alpha = 1, \beta = 6, \gamma = 3$

Now, required line having direction cosine \left(l, m, n\right)$, where given$m = \cos \left(\dfrac{- 2\pi}{3}\right) = \dfrac{- 1}{2} \text{ and } n = \cos \left(\dfrac{3\pi}{4}\right) = \dfrac{- 1}{\sqrt{2}}$</p><p>Also,${l}^{2} + {m}^{2} + {n}^{2} = 1$</p><p>$⇒ {l}^{2} + {\left(- \dfrac{1}{2}\right)}^{2} + {\left(- \dfrac{1}{\sqrt{2}}\right)}^{2} = 1$</p><p>$⇒ {l}^{2} = \dfrac{1}{4}$</p><p>$⇒ l = \dfrac{1}{2}$</p><p>[Line make acute angle with$x -$axis]</p><p>So, the equation of line passing through \left(1, 6, 3\right)$ will be given by,

\overset{\to}{r} = ( \overset{^}{i} + 6\overset{^}{j} + 3\overset{^}{k} ) + \mu\left(\dfrac{1}{2}\overset{^}{i} - \dfrac{1}{2}\overset{^}{j} - \dfrac{1}{\sqrt{2}}\overset{^}{k}\right)

Option \left(3\right)$\left(3, 4, 3 - 2\sqrt{2}\right)$ satisfying for $\mu = 4$