The foot of the perpendicular drawn from the point $(1,6,3)$ to the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$ is

The given line is $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$

This can be rewritten as $\frac{x-0}{1} = \frac{y-1}{2} = \frac{z-2}{3}$

The line passes through the point $(0, 1, 2)$ with direction ratios $(1, 2, 3)$.

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Let the common ratio be $t$. Then:

$\frac{x}{1} = t$, $\frac{y-1}{2} = t$, $\frac{z-2}{3} = t$

Solving for $x, y, z$:

$x = t$

$y = 1 + 2t$

$z = 2 + 3t$

Any point on this line can be written as $(t, 1+2t, 2+3t)$.

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Let the foot of perpendicular be $P = (t, 1+2t, 2+3t)$ and the given point be $A = (1, 6, 3)$.

The vector $\vec{AP} = (t-1, 1+2t-6, 2+3t-3)$

$\vec{AP} = (t-1, 2t-5, 3t-1)$

For $P$ to be the foot of perpendicular, $\vec{AP}$ must be perpendicular to the direction vector of the line.

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The direction vector of the line is $(1, 2, 3)$.

For perpendicular vectors, their dot product equals zero:

$(t-1, 2t-5, 3t-1) \cdot (1, 2, 3) = 0$

$(t-1)(1) + (2t-5)(2) + (3t-1)(3) = 0$

$t - 1 + 4t - 10 + 9t - 3 = 0$

$14t - 14 = 0$

$t = 1$

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Substituting $t = 1$ into $(t, 1+2t, 2+3t)$:

$x = 1$

$y = 1 + 2(1) = 3$

$z = 2 + 3(1) = 5$

Therefore, the foot of perpendicular is $(1, 3, 5)$.