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Question: If the points \((1,1,k)\) and \(( - 3,0,1)\) be equidistant from the plane \(3x + 4y - 12z + 13 = 0\...

If the points (1,1,k)(1,1,k) and (−3,0,1)( - 3,0,1) be equidistant from the plane 3x+4y−12z+13=03x + 4y - 12z + 13 = 0,then k =

A

0

B

1

C

2

D

None of these

Answer

1

Explanation

Solution

According to question, ∣3+4−12k+13∣=∣−9−12+13∣| 3 + 4 - 12 k + 13 | = | - 9 - 12 + 13 |

∴3+4−12k+13=8⇒k=1\therefore 3 + 4 - 12 k + 13 = 8 \Rightarrow k = 1.