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(B) Let the equation of line $l$ be $\frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k$. Since $l$ is perpendicular to $l_1$ and $l_2$,its direction ratios

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$(B, D)$

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(B) Let the equation of line $l$ be $\frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k$. Since $l$ is perpendicular to $l_1$ and $l_2$,its direction ratios $(a, b, c)$ must satisfy the dot product with the direction vectors of $l_1$ and $l_2$ being zero.Direction of $l_1$ is $\vec{v_1} = (1, 2, 2)$ and direction of $l_2$ is $\vec{v_2} = (2, 2, 1)$.Thus,$a + 2b + 2c = 0$ and $2a + 2b + c = 0$.Solving these,we get $\frac{a}{2-4} = \frac{b}{4-1} = \frac{c}{2-4} \Rightarrow \frac{a}{-2} = \frac{b}{3} = \frac{c}{-2}$.So,line $l$ is $\frac{x}{-2} = \frac{y}{3} = \frac{z}{-2} = k_1$. The point of intersection $P$ of $l$ and $l_1$ is found by setting $(-2k_1, 3k_1, -2k_1) = (3+t, -1+2t, 4+2t)$.Solving $3+t = -2k_1$ and $-1+2t = 3k_1$ and $4+2t = -2k_1$,we find $k_1 = -1$,so $P = (2, -3, 2)$.Let a point $Q$ on $l_2$ be $(3+2s, 3+2s, 2+s)$. The distance $PQ = \sqrt{17}$ implies:$(3+2s-2)^2 + (3+2s+3)^2 + (2+s-2)^2 = 17$$(1+2s)^2 + (6+2s)^2 + s^2 = 17$$1 + 4s + 4s^2 + 36 + 24s + 4s^2 + s^2 = 17$$9s^2 + 28s + 37 = 17 \Rightarrow 9s^2 + 28s + 20 = 0$$(9s + 10)(s + 2) = 0 \Rightarrow s = -2, s = -\frac{10}{9}$.For $s = -2$,$Q = (-1, -1, 0)$.For $s = -\frac{10}{9}$,$Q = (3 - \frac{20}{9}, 3 - \frac{20}{9}, 2 - \frac{10}{9}) = (\frac{7}{9}, \frac{7}{9}, \frac{8}{9})$.Thus,the points are $(-1, -1, 0)$ and $(\frac{7}{9}, \frac{7}{9}, \frac{8}{9})$,which corresponds to option $(B, D)$.

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