$$ \begin{aligned} & \ell_1: r=(3+t) \hat{i}+(-1+2 t) \hat{j}+(4+2 t) \hat{k} \\ & \ell_2: r=(3+2 s) \hat{i}+(3+2 s) \hat{j}+(4+s) \hat{k} \end{aligned} $$ DR of $\square_1 \equiv(1,2,2)$
DR of $\square_2 \equiv(2,2,1)$
DR of □ (line $\perp$ to $\square_1$ \& $\square_2$ )
$$ \begin{aligned} & =(-2,3,-2) \\ & \therefore \quad \ell: \mathrm{r}=-2 \mu \hat{\mathrm{i}}+3 \mu \hat{\mathrm{j}}-2 \mu \hat{\mathrm{k}} \end{aligned} $$ for intersection of $\square$ \& $\square_1$ $$ \begin{aligned} & 3+t=-2 \mu \\ & -1+2 t=3 \mu \\ & 4+2 t=-2 \mu \\ & \Rightarrow t=-1 \& \lambda=-1 \end{aligned} $$ ∴ Point of intersection $P \equiv(2,-3,2)$
Let point on $\square_2$ be $\mathrm{Q}(3+2 \mathrm{~s}, 3+2 \mathrm{~s}, 2+\mathrm{s})$
Given $\mathrm{PQ}=\sqrt{17}$
$$ \begin{aligned} & \Rightarrow(P Q)^2=17 \\ & \Rightarrow(2 s+1)^2+(6+2 s)^2+(s)^2=17 \\ & \Rightarrow 9 s^2+28 s+20=0 \\ & \Rightarrow \quad s=-2,-\frac{10}{9} \\ & s \neq-2 \text { as point lies on } 1^{\text {st }} \text { octant. } \\ & \quad \quad \quad \quad a=3+2\left(-\frac{10}{9}\right)=\frac{7}{9} \\ & b=3+2\left(-\frac{10}{9}\right)=\frac{7}{9} \\ & c=2+\left(-\frac{10}{9}\right)=\frac{8}{9} \\ & \quad \quad 18(a+b+c)=18\left(\frac{22}{9}\right)=44 \end{aligned} $$