The vector equation of line passing through (2, -1, 3) and perpendicular to the lines

$\frac{x-2}{3} = \frac{y-1}{1} = \frac{z+2}{2}$ and $\frac{x+3}{-4} = \frac{y-5}{-3} = \frac{z+1}{2}$ is

(Here $\lambda$ is a parameter)

The line must pass through $(2, -1, 3)$ and be perpendicular to the two given lines.

For the line $\frac{x-2}{3} = \frac{y-1}{1} = \frac{z+2}{2}$, the direction vector is:

$\vec{b_1} = 3\hat{i} + 1\hat{j} + 2\hat{k}$

For the line $\frac{x+3}{-4} = \frac{y-5}{-3} = \frac{z+1}{2}$, the direction vector is:

$\vec{b_2} = -4\hat{i} - 3\hat{j} + 2\hat{k}$

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The direction vector of the required line is perpendicular to both given lines, so it can be found using the cross product:

$\vec{b} = \vec{b_1} \times \vec{b_2}$

$\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ -4 & -3 & 2 \end{vmatrix}$

$\vec{b} = \hat{i}(1 \cdot 2 - 2 \cdot (-3)) - \hat{j}(3 \cdot 2 - 2 \cdot (-4)) + \hat{k}(3 \cdot (-3) - 1 \cdot (-4))$

$\vec{b} = \hat{i}(2 + 6) - \hat{j}(6 + 8) + \hat{k}(-9 + 4)$

$\vec{b} = 8\hat{i} - 14\hat{j} - 5\hat{k}$

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The vector equation of a line passing through a point with position vector $\vec{a}$ and having direction vector $\vec{b}$ is:

$\vec{r} = \vec{a} + \lambda\vec{b}$

The position vector of point $(2, -1, 3)$ is $2\hat{i} - \hat{j} + 3\hat{k}$

The direction vector is $8\hat{i} - 14\hat{j} - 5\hat{k}$

Therefore, the vector equation is:

$\vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda(8\hat{i} - 14\hat{j} - 5\hat{k})$