Let the line of the shortest distance between the lines L_1: =( i+2 j+3 k)+λ ( i- j+ k) and L_2: =(4 i+5 j+6 k)+μ ( i+ j- k) intersect L_1 and L_2 at…

Let point on ${L}_{1}$L1​ be $P((1+\lambda )\hat{i}+(2-\lambda )\hat{j}+(3+\lambda )\hat{k})$P((1+λ)i^+(2−λ)j^​+(3+λ)k^) and point on ${L}_{2}$L2​ be $Q((4+\mu )\hat{i}+(5+\mu )\hat{j}+(6-\mu )\hat{k})$Q((4+μ)i^+(5+μ)j^​+(6−μ)k^).

$$\Rightarrow \vec{PQ}=(\lambda -\mu -3)\hat{i}+(-\lambda -\mu -3)\hat{j}+(\lambda +\mu -3)\hat{k}$$⇒PQ​=(λ−μ−3)i^+(−λ−μ−3)j^​+(λ+μ−3)k^

Now, $\vec{PQ}\perp {L}_{1},{L}_{2}$PQ​⊥L1​,L2​

$$\Rightarrow (\lambda -\mu -3)+(\lambda +\mu +3)+(\lambda +\mu -3)=0$$⇒(λ−μ−3)+(λ+μ+3)+(λ+μ−3)=0

$$\Rightarrow 3\lambda +\mu =3...(i)$$⇒3λ+μ=3...(i)

Also, $(\lambda -\mu -3)+(-\lambda -\mu -3)+(-\lambda -\mu +3)=0$(λ−μ−3)+(−λ−μ−3)+(−λ−μ+3)=0

$$\Rightarrow -\lambda -3\mu =3...(ii)$$⇒−λ−3μ=3...(ii)

Now, solving above equations we get,

$$\Rightarrow 3\lambda +\mu -3\lambda -9\mu =3+9$$⇒3λ+μ−3λ−9μ=3+9

$$\Rightarrow -8\mu =12$$⇒−8μ=12

$$\Rightarrow \mu =\frac{-3}{2}$$⇒μ=2−3​

$$\Rightarrow \lambda =\frac{3}{2}$$⇒λ=23​

$$\Rightarrow P\equiv (\frac{5}{2},\frac{1}{2},\frac{9}{2}),Q\equiv (\frac{5}{2},\frac{7}{2},\frac{15}{2})$$⇒P≡(25​,21​,29​),Q≡(25​,27​,215​)

So, mid-point of $PQ$PQ is $(\frac{5}{2},2,6)$(25​,2,6)

$$\Rightarrow 2(\alpha +\beta +\gamma )=2(\frac{5}{2}+2+6)$$⇒2(α+β+γ)=2(25​+2+6)

$$\Rightarrow 2(\alpha +\beta +\gamma )=21$$⇒2(α+β+γ)=21