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Mathematics
Problem
The distance of the point
(
1
,
0
,
2
)
(1, 0, 2)
(
1
,
0
,
2
)
from the point of intersection of the line
x
−
2
3
=
y
+
1
4
=
z
−
2
12
\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}
3
x
−
2
=
4
y
+
1
=
12
z
−
2
and the plane
x
−
y
+
z
=
16
x - y + z = 16
x
−
y
+
z
=
16
, is a.
2
14
2\sqrt{14}
2
14
b. 8 c.
3
21
3\sqrt{21}
3
21
d. 13
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