1.

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

Answer» Correct Answer - B
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))`


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