JEE Math Practice Question
The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line $${x \over 2} = {y \over 3} = {z \over { - 6}}$$ is :
- A.7
- B.1
- C.$${1 \over 7}$$
- D.$${7 \over 5}$$
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