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Question

The distance of the point (1, 0, 2) from the point of intersection of the line
x23=y+14=z212
and the plane xy+z=16, is

A
214
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B
8
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C
321
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D
13
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Solution

The correct option is D 13
We will take the point of intersection as the general point on the line.
(3λ+2,4λ1,12λ+2)
Now we will substitute this in the equation of the plane.
3λ+24λ+1+12λ+2=16
11λ=11 λ=1
(5, 3, 14) is the point of intersection.
Distance = 16+9+144=169=13

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