The distance of the point (1, 0, 2) from the point of intersection of the line x−23=y+14=z−212
and the plane x−y+z=16, is
A
2√14
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B
8
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C
3√21
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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+2,4λ−1,12λ+2)
Now we will substitute this in the equation of the plane. 3λ+2−4λ+1+12λ+2=16 11λ=11⇒λ=1
(5, 3, 14) is the point of intersection.
Distance = √16+9+144=√169=13