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

Let the foot point of the perpendicular drawn
from the point \(P(3,-1,1)\) on the straight line be \(L\).
\(\therefore\) Hence, \(L\) lies on the straight line
\(\therefore L(2 t, 2+3 t, 3+4 t)\) [where, \(t\) is arbitrary
constant]
\(\therefore\) The direction ratios of \(P L\) are \((2 t-3,2+3 t+1\),
Again, the direction ratios of the straight line.
\(\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) are \((2,3,4)\).
Since, \(P L\) is perpendicular on the straight line.
Then, \((2 t-3) \cdot 2+(3 t+3) \cdot 3+(4 t-8) \cdot 4=0\)
\(\begin{aligned} & 4 t-6+9 t+9+16 t-32=0 \\ & 29 t=29\end{aligned}\)
\(\because \quad t=1\)
Hence, \(L(2,5,7)\)
\(\therefore \quad(P L)=\sqrt{(2-3)^2+(5+1)^2+(7-11)^2}=\sqrt{53}\)