The distance of the point (1,0,2) from the point of intersection of the line x/3 = y+1/4 = z/12 and the plane x-y+z=16, is

Let x/3 = y+1/4 = z/12 = t
Any point on the line can be written in the parametric form as (3t, 4t-1, 12t).
To find the point of intersection, let us substitute the point in the equation of the plane.
⇒ 3t - (4t-1) + 12t = 16
⇒ 11t = 15
⇒ t = 15/11
Hence, the point of intersection is (45/11, 55/11, 180/11)
The distance of (45/11, 59/11, 180/11) from (1,0,2) is given by
√[(45/11 - 1)² + (59/11)² + (180/11 -2)²]
= √[(34/11)² + (59/11)² + (158/11)²]
= √(1156 + 3481 + 24964)/11
= √29601/11
= 172.04/11
≈ 15.64

Let's reconsider the parametric form. Let x/3 = y+1/4 = z/12 = t. Then x = 3t, y = 4t - 1, z = 12t.
Substituting into the plane equation x - y + z = 16, we get:
3t - (4t - 1) + 12t = 16
11t + 1 = 16
11t = 15
t = 15/11
The point of intersection is (45/11, 59/11, 180/11).
The distance between (1, 0, 2) and (45/11, 59/11, 180/11) is:
√[(45/11 - 1)² + (59/11 - 0)² + (180/11 - 2)²]
= √[(34/11)² + (59/11)² + (158/11)²]
= √(1156 + 3481 + 24964)/11
= √29601/11 ≈ 15.64

There seems to be a mistake in the original solution or question. Let's use the corrected parametric form:
Let x/3 = (y+1)/4 = (z)/12 = k
Then x = 3k, y = 4k - 1, z = 12k
Substituting into x - y + z = 16:
3k - (4k - 1) + 12k = 16
11k = 15
k = 15/11
Intersection point: (45/11, 59/11, 180/11)
Distance = √[(45/11 - 1)² + (59/11)² + (180/11 - 2)²] ≈ 15.64. None of the options match.