Let the image of the point (1,0,7) in the line x 1= y-1 2= z-2 3 be the point (α ,β ,γ ). Then which one of the following points lies on the line…

Given:

$${L}_{1}\equiv \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda$$L1​≡1x​=2y−1​=3z−2​=λ

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

$$\Rightarrow \vec{\mathrm{PM}}=(\lambda -1)\hat{i}+(1+2\lambda )\hat{j}+(3\lambda -5)\hat{k}$$⇒PM=(λ−1)i^+(1+2λ)j^​+(3λ−5)k^

So, $\vec{\mathrm{PM}}$PM is perpendicular to line ${L}_{1}$L1​

$$\Rightarrow \vec{\mathrm{PM}}\cdot \vec{b}=0$$⇒PM⋅b=0

$$\Rightarrow \vec{\mathrm{PM}}.\vec{b}=[(\lambda -1)\hat{i}+(1+2\lambda )\hat{j}+(3\lambda -5)\hat{k}].(\hat{i}+2\hat{j}+3\hat{k})$$⇒PM.b=[(λ−1)i^+(1+2λ)j^​+(3λ−5)k^].(i^+2j^​+3k^)

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

$$\Rightarrow 14\lambda =14$$⇒14λ=14

$$\Rightarrow \lambda =1$$⇒λ=1

$$\Rightarrow M=(1,3,5)$$⇒M=(1,3,5)

Now, $M$M is midpoint of $\vec{P}$P and $\vec{Q}$Q​

$$\Rightarrow (1,3,5)\equiv (\frac{1+\alpha }{2},\frac{0+\beta }{2},\frac{7+\gamma }{2})$$⇒(1,3,5)≡(21+α​,20+β​,27+γ​)

$$\Rightarrow \frac{1+\alpha }{2}=1,\frac{\beta }{2}=3,\frac{7+\gamma }{2}=5$$⇒21+α​=1,2β​=3,27+γ​=5

$$\Rightarrow \alpha =1,\beta =6,\gamma =3$$⇒α=1,β=6,γ=3

Now, required line having direction cosine $(l,m,n)$(l,m,n), where given m=\mathrm{cos}(\frac{-2\pi }{3})=\frac{-1}{2}&n=\mathrm{cos}(\frac{3\pi }{4})=\frac{-1}{\sqrt{2}}

Also, ${l}^{2}+{m}^{2}+{n}^{2}=1$l2+m2+n2=1

$$\Rightarrow {l}^{2}+{(-\frac{1}{2})}^{2}+{(-\frac{1}{\sqrt{2}})}^{2}=1$$⇒l2+(−21​)2+(−2​1​)2=1

$$\Rightarrow {l}^{2}=\frac{1}{4}$$⇒l2=41​

$$\Rightarrow l=\frac{1}{2}$$⇒l=21​

[Line make acute angle with $x-$x−axis]

So, the equation of line passing through $(1,6,3)$(1,6,3) will be given by,

$$\vec{r}=(\hat{i}+6\hat{j}+3\hat{k})+\mu (\frac{1}{2}\hat{i}-\frac{1}{2}\hat{j}-\frac{1}{\sqrt{2}}\hat{k})$$r=(i^+6j^​+3k^)+μ(21​i^−21​j^​−2​1​k^)

Option $(3)$(3) $(3,4,3-2\sqrt{2})$(3,4,3−22​) satisfying for $\mu =4$μ=4