$${l_1}:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$$

$${l_1}:{{x - 3} \over 1} = {{y + 1} \over 2} = {{z - 4} \over 2} \Rightarrow $$ D.R. of $${l_1} = 1,2,2$$

$${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$$

$${l_2}:{{x - 3} \over 2} = {{y - 3} \over 2} = {{z - 2} \over 1} \Rightarrow $$ D.R. of $${l_2} = 2,2,1$$

D.R. of l is $$ \bot $$ to l₁ & k₂

$$ \therefore $$ D.R. of $$l\,||\,({l_1} \times {l_2}) \Rightarrow ( - 2,3 - 2)$$

$$ \therefore $$ Equation of $$l:{x \over 2} = {y \over { - 3}} = {z \over 2}$$

Solving l & l₁

$$(2\lambda , - 3\lambda ,2\lambda ) = (\mu + 3,2\mu - 1,2\mu + \mu )$$

$$ \Rightarrow 2\lambda = \mu + 3$$

$$ - 3\lambda = 2\mu - 1$$

$$2\lambda = 2\mu + 4$$

$$ \Rightarrow \mu + 3 = 2\mu + 4$$

$$\mu = - 1$$

$$\lambda = 1$$

$$P(2, - 3,2)$$ {intersection point}

Let, $$Q(2v + 3,2v + 3,v + 2)$$ be point on l₂

Now, $$PQ = \sqrt {{{(2v + 3 - 2)}^2} + {{(2v + 3 + 3)}^2} + {{(v + 2 - 2)}^2}} = \sqrt {17} $$

$$ \Rightarrow {(2v + 1)^2} + {(2v + 6)^2} + {(v)^2} = 17$$

$$ \Rightarrow 9{v^2} + 28v + 36 + 1 - 17 = 0$$

$$ \Rightarrow 9{v^2} + 28v + 20 = 0$$

$$ \Rightarrow 9{v^2} + 18v + 10v + 20 = 0$$

$$ \Rightarrow (9v + 10)(v + 2) = 0$$

$$ \Rightarrow v = - 2$$ (rejected), $$ - {{10} \over 9}$$ (accepted)

$$Q\left( {3 - {{20} \over 9},3 - {{20} \over 9},2 - {{10} \over 9}} \right)$$

$$\left( {{7 \over 9},{7 \over 9},{8 \over 9}} \right)$$

$$ \therefore $$ $$18(a + b + c)$$

$$ = 18\left( {{7 \over 9},{7 \over 9},{8 \over 9}} \right)$$

$$ = 44$$