1.

Find the distance of thepoint `(-1,-5,-10)`from the point ofintersection of the line `(x-2)/3=(y+1)/4=(z-2)/(12)`and plane `x-y+z=5.`

Answer» Any point on the line `(x-2)/(3)=(y+1)/(4)=(z-2)/(12)= lamda` is
`(3lamda+2, 4lamda-1, 12lamda+2)`.
This lies on `x-y+z=5`.
If `3lamda+2-4lamda+1+12lamda+2=5 or lamda=0`, then the point is `(2, -1, 2)`.
Its distance from `(-1, -5, -10)` is
`" "sqrt((2+1)^(2)+(-1+5)^(2)+(2+10)^(2))`
`" "=sqrt(9+16+144)=13`


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