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Let the general point on the first line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} = \lambda$ be $P(\lambda, 2\lambda+1, 3\lambda+2)$.\nLet the general point on the second line $\frac{x-1}{0}=\frac{y}{-3}=\frac{z-7}{2} = \mu$ be $Q(1, -3\mu, 2\mu+7)$.
For the lines to intersect, the coordinates of $P$ and $Q$ must be equal for some $\lambda$ and $\mu$.\nSo, we have:\n$\lambda = 1$\n$2\lambda + 1 = -3\mu$\n$3\lambda + 2 = 2\mu + 7$\nSubstituting $\lambda = 1$ into the second equation:\n$2(1) + 1 = -3\mu \implies 3 = -3\mu \implies \mu = -1$\nSubstituting $\lambda = 1$ into the third equation:\n$3(1) + 2 = 2\mu + 7 \implies 5 = 2\mu + 7 \implies 2\mu = -2 \implies \mu = -1$\nSince both equations give $\mu = -1$, the lines intersect at a point.\nSubstituting $\lambda = 1$ into the coordinates of $P$, we get the point of intersection as $(1, 3, 5)$.
The direction vector of the first line is $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.\nThe direction vector of the second line is $\vec{b} = 0\hat{i} - 3\hat{j} + 2\hat{k}$.
The required line is perpendicular to both given lines. Therefore, its direction vector $\vec{c}$ is the cross product of $\vec{a}$ and $\vec{b}$.\n$\vec{c} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & -3 & 2 \end{vmatrix} = (4 - (-9))\hat{i} - (2 - 0)\hat{j} + (-3 - 0)\hat{k} = 13\hat{i} - 2\hat{j} - 3\hat{k}$
The equation of the line passing through the point $(1, 3, 5)$ and having direction vector $13\hat{i} - 2\hat{j} - 3\hat{k}$ is given by:\n$\frac{x - 1}{13} = \frac{y - 3}{-2} = \frac{z - 5}{-3}$
\r\n Final Answer: $\frac{x - 1}{13} = \frac{y - 3}{-2} = \frac{z - 5}{-3}$<\/span>\r\n <\/p>\r\n <\/div>\r\n <\/div>
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