The distance of the point `(1,3,-7)` from the plane passing through the point `(1,-1,-1),` having normal perpendicular to both the lines `(x-1)/1=(y+2)/(-2)=(z-4)/3a n d(x-2)/2=(y+1)/(-1)=(z+7)/(-1)i s:` `5/(sqrt(83))` (2) `(10)/(sqrt(74))` (3) `(20)/(sqrt(74))` (4) `(10)/(sqrt(83))`

Explanation

Let `vecn` is the vector normal to the plane.

Then, as `vecn` is perpendicular to given two lines.

`:. vec n = |[hati,hatj,hatk],[1,-2,3],[2,-1,-1]|`

`=> vec n = 5hati + 7hat j+3hat k`

Now, as `vec n` is normal to the plane passing through point `(1,-1,1)`,

`:. (xhati+yhatj+zhatk - hati+hatj+hatk).(5hati + 7hat j+3hat k) = 0`

`=>5x-5+7y+7+3z+3 = 0`

`=>5x+7y+3z+5 = 0`

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