If (2, 3, 5) is one end of a diameter of the sphere `x^2+""y^2+""z^2-6x-12 y-2z""+""20""=""0` , then the coordinates of the other end of the diameter are (1) `(4,""9,-3)` (2) `(4,-3,""3)` (3) `(4,""3,""5)` (4) `(4,""3,-3)`

Options

A. `(x-1)/(2)=(y+2)/(-1)=(z-2)/(2)`
B. `(x+(1//2))/(1)=(y-1)/(-2)=(z-(1//2))/(1)`
C. `(x)/(1)=(y)/(-2)=(z-1)/(1)`
D. `(x+1)/(1)=(y-2)/(-2)=(z=0)/(1)`

Answer

b, c, d

Explanation

`" "x+y+z-1=0`
`" "4x+y-2z+2=0`
Therefore, the line is along the vector
`" "(hati+hatj+hatk)xx(4hati+hatj-2hatk)=3hati-6hatj+3hatk`
Let `z=k`. Then `x=k-1 and y=2-2k`
Therefore, `(k-1, 2-2k, k)` is any point on the line
Hence, `(-1, 2, 0), (0, 0, 1) and (-1//2, 1, 1//2)` are the points on the line.