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)/(3) and (x-2)/(2)=(y+1)/(-1)=(z+7)/(-1)` is

Options

A. `20/(sqrt(74))`
B. `10/(sqrt(83))`
C. `10/(sqrt(74))`
D. `5/(sqrt(83))`

Answer

B

Explanation

Let the equation of the plane through the point (1,-1,-1) be
`a(x-1)+b(y+1)+c(z+1)=0`……………..i
The direction ratios of the normal to the plane proportional to a,b,c. The normal is perpendicular to the lines
`(x-1)/1=(y+2)/(-2)=(z-4)/3` and `(x-2)/2=(y+1)/(-1)=(z+7)/(-1)`
`:.a-2b+3c=0` and `2a-b-c=0`
Using cross multiplication, we obtain
`5(x-1)+7(y+1)+3(z+1)-0` or `5x+7y+3z+5=0` ................ii
The distance `d` of the point (1,3,-7) from plane ii is
`d=|(5+21-21+5)/(sqrt(25+49+9))|=10/(sqrt(83))`