Given line is \(\frac{x}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z-3}{4}\) . 

The direction ratios of the line are 2, 3 and 4. 

Now,  \(\frac{x}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z-3}{4}\) = \(\lambda\) (Let)

⇒ x = 2\(\lambda\), = 3\(\lambda\) + 2 and = 4\(\lambda\) + 3.

Hence, the points on the line \(\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4}\) is of the form Q (2\(\lambda\), 3\(\lambda\) + 2, 4\(\lambda\) + 3).

The direction ratios of the line joining points (3, −1, 11) and (2\(\lambda\), 3\(\lambda\) + 2, 4\(\lambda\) + 3) are

2\(\lambda\) − 3, 3\(\lambda\) + 2 − (−1) and 4\(\lambda\) + 3 − 11 = 2\(\lambda\) − 3, 3\(\lambda\) + 3 and 4\(\lambda\) − 8.

Let the foot of the perpendicular is Q (2\(\lambda\), 3\(\lambda\) + 2, 4\(\lambda\) + 3).

Let line PQ is perpendicular to the given line \(\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4}.\)

Therefore, 2(2\(\lambda\) - 3)+3(3\(\lambda\) + 3) + 4(4\(\lambda\) - 8) = 0.

(Because, the sum of product of direction ratios of two perpendicular lines is equal to zero.) 

⇒ 4\(\lambda\) − 6 + 9\(\lambda\) + 9 + 16\(\lambda\) − 32 = 0

⇒ 29\(\lambda\) − 29 = 0 ⇒ \(\lambda\) = \(\frac{29}{29}= 1\).

Now, putting the value of = 1 in the equation of points on the given line. 

Therefore, the foot of the perpendicular is Q (2\(\lambda\), 3\(\lambda\) + 2, 4\(\lambda\) + 3) = Q(2, 5, 7).

The equation of line PQ is \(\frac{x-3}{2-3} = \frac{y-(-1)}{5-(-1)}\) = \(\frac{z-11}{7-11} ⇒ \frac{x-3}{-1}\) = \(\frac{y+1}{6}\) = \(\frac{z-11}{-4}\)

The equation of the perpendicular from the point (3, −1, 11) to the line \(\frac{x}{2} = \frac{y-2}{3}\) = \(\frac{z-3}{4}\) is 

\(\frac{x-3}{-1} = \frac{y+1}{6} = \frac{z-11}{-4}\).

Now, length of perpendicular is = \(\sqrt{(2-3)^2 + (5-(-1))^2+ (7-11)^2}\)

= \(\sqrt{(-1)^2 + 6^2 + (-4)^2}\) = \(\sqrt{1+36+16}\) = \(\sqrt{53}\) units.