Find the distance between the line `(x+1)/(-3)=(y-3)/2=(z-2)/1` and the plane `x+y+z+3=0.`

Explanation

The line is along the vector `veca= - 3hati+2hatj+hatk ` and plane is normal to the vector `vecb= hati+hatk+hatk`.
Since `veca*vecb = 0` , the line is parallel to the plane.
Hence, the distance between the line and the plane is the distance of point `(-1, 3, 2)` from the plane,
`" "(|-1+3+2+3|)/(sqrt(1+1+1))= (7)/(sqrt(3))`