The correct option is 34/√1045 unit.

To find the shortest distance between two skew lines in three-dimensional space, we can analyze the equations of the lines from the provided images:
Line 1:

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

From this equation, we identify a point on Line 1 as:
$$\overrightarrow{a_1}=\hat{i}-\hat{j}+\hat{k}$$
And its direction vector as:
$$\overrightarrow{n_1}=2\hat{i}+4\hat{j}+3\hat{k}$$

Line 2:
The given equation is:

$$\frac{2x-1}{5}=\frac{y-2}{3}=\frac{z}{6}$$

To write Line 2 in standard symmetric form, we divide the numerator and denominator of the first term by 2:
$$\frac{x-\frac{1}{2}}{\frac{5}{2}}=\frac{y-2}{3}=\frac{z}{6}$$

From this standard form, we identify a point on Line 2 as:
$$\overrightarrow{a_2}=\frac{1}{2}$$
And its direction vector as:
$$\overrightarrow{n_2}=\frac{5}{2}\hat{i}+3\hat{j}+6\hat{k}$$

Step 1: Calculate the difference vector between the two points
Subtracting the position vector of the point on the second line from the first:
$$\overrightarrow{a_1}-\overrightarrow{a_2}=\left(1-\frac{1}{2}\right)\hat{i}+\left(-1-2\right)\hat{j}+\left(1-0\right)\hat{k}$$
$$\overrightarrow{a_1}-\overrightarrow{a_2}=\frac{1}{2}\hat{i}-3\hat{j}+\hat{k}$$

Step 2: Find the cross product of the direction vectors
The vector normal to both lines is given by the cross product:
$$\overrightarrow{n_1}∕\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 4& 3\\ \frac{5}{2}& 3& 6\end{array}\right|$$
Expanding this determinant along the first row:
$$\overrightarrow{n_1}∕\times \overrightarrow{n_2}=\hat{i}\left(4\times 6-3\times 3\right)-\hat{j}\left(2\times 6-3\times \frac{5}{2}\right)+\hat{k}\left(2\times 3-4\times \frac{5}{2}\right)$$
$$\overrightarrow{n_1}∕\times \overrightarrow{n_2}=\hat{i}\left(24-9\right)-\hat{j}\left(12-\frac{15}{2}\right)+\hat{k}\left(6-10\right)$$
$$\overrightarrow{n_1}∕\times \overrightarrow{n_2}=15\hat{i}-\frac{9}{2}\hat{j}-4\hat{k}$$

Step 3: Calculate the magnitude of the cross product vector
$$|\overrightarrow{n_1}∕\times \overrightarrow{n_2}|=\sqrt{{15}^2+{(-\frac{9}{2})}^2+{(-4)}^2}$$
$$|\overrightarrow{n_1}∕\times \overrightarrow{n_2}|=\sqrt{225+\frac{81}{4}+16}=\sqrt{241+\frac{81}{4}}$$
Finding the common denominator:
$$|\overrightarrow{n_1}∕\times \overrightarrow{n_2}|=\sqrt{\frac{964+81}{4}}=\sqrt{\frac{1045}{4}}=\frac{\sqrt{1045}}{2}$$

Step 4: Find the dot product of the difference vector and the cross product vector
$$\left(\overrightarrow{a_1}-\overrightarrow{a_2}\right)\cdot \left(\overrightarrow{n_1}∕\times \overrightarrow{n_2}\right)=\left(\frac{1}{2}\right)\left(15\right)+\left(-3\right)\left(-\frac{9}{2}\right)+\left(1\right)\left(-4\right)$$
$$=\frac{15}{2}+\frac{27}{2}-4=\frac{42}{2}-4=21-4=17$$

Step 5: Apply the shortest distance formula
The shortest distance $d$ is given by:
$$d=\left|\frac{\left(\overrightarrow{a_1}-\overrightarrow{a_2}\right)\cdot \left(\overrightarrow{n_1}∕\times \overrightarrow{n_2}\right)}{|\overrightarrow{n_1}∕\times \overrightarrow{n_2}|}\right|$$
Substituting the calculated values:
$$d=\frac{17}{\frac{\sqrt{1045}}{2}}=\frac{34}{\sqrt{1045}}\text{ unit}$$