Explanation
\begin{aligned}
& \mathrm{Here},\vec{a}=4\hat{i}-\hat{j}+0\hat{k};\vec{b}=\lambda\hat{i}-\hat{j}+2\hat{k} \\
& \vec{p}=\hat{i}+2\hat{j}-3\hat{k}\mathrm{and}\vec{q}=2\hat{i}+4\hat{j}-5\hat{k} \\
& \mathrm{So},\vec{b}-\vec{a}=(\lambda-4)\hat{i}+2\hat{k} \\
& \vec{v}\times\vec{q}=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & -3 \\
2 & 4 & -5
\end{vmatrix}=\hat{i}(-10+12)-\hat{j}(-5+6)+\hat{k}(4-4) \\
& =2\hat{i}-\hat{j}+0\hat{k}
\end{aligned}
p×q=4+1=5
So, shortest distance =56
=5(λ−4)j^+2k^)(2i^−j^)
=5(λ−4)2i^+(λ−4)(−1)j^+2k^
=52λ−8=6⇒∣λ−4∣=3
⇒λ=7,1
So, required sum =7+1=8
