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(D) Let the given line be $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2} = r$. Any point $P$ on $L_1$ is given by $P(2r+1, 3r-1, -2r+1)$. The direct

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$\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}$

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(D) Let the given line be $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2} = r$. Any point $P$ on $L_1$ is given by $P(2r+1, 3r-1, -2r+1)$. The direction vector of $L_1$ is $\vec{v_1} = 2\hat{i} + 3\hat{j} - 2\hat{k}$. The vector $\vec{QP}$ is $(2r+1-0)\hat{i} + (3r-1-1)\hat{j} + (-2r+1-2)\hat{k} = (2r+1)\hat{i} + (3r-2)\hat{j} + (-2r-1)\hat{k}$. Since the required line is perpendicular to $L_1$,$\vec{QP} \cdot \vec{v_1} = 0$. $2(2r+1) + 3(3r-2) - 2(-2r-1) = 0$. $4r + 2 + 9r - 6 + 4r + 2 = 0$. $17r - 2 = 0 \Rightarrow r = \frac{2}{17}$. The direction ratios of the line $QP$ are proportional to the components of $\vec{QP}$ at $r = \frac{2}{17}$. $\vec{QP} = (2(\frac{2}{17})+1)\hat{i} + (3(\frac{2}{17})-2)\hat{j} + (-2(\frac{2}{17})-1)\hat{k} = \frac{21}{17}\hat{i} - \frac{28}{17}\hat{j} - \frac{21}{17}\hat{k}$. Dividing by $\frac{7}{17}$,the direction ratios are $(3, -4, -3)$. Thus,the equation of the line passing through $(0,1,2)$ with direction ratios $(3, -4, -3)$ is $\frac{x-0}{3} = \frac{y-1}{-4} = \frac{z-2}{-3}$,which is equivalent to $\frac{x}{-3} = \frac{y-1}{4} = \frac{z-2}{3}$.

The equation of the line passing through the point $Q(0,1,2)$ and perpendicular to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}$ is

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