find the distance from the line x=2+t, y=1+t, z= -(1/2)-(1/2)t to the plane x+2y+6z =10

my first attempt ::

by setting t = 0, we get a point on the line, say P :<2,1,-1/2>. the point on the plane, say, S we get if we set y=0, z=0 ==> x= 0. thus the point S on the plane is <10,0,0>.

thus the vector PS = S - P =8i-j+1/2k

vector normal to the plane, n = i+2j+6k

so the distance point P to the plane = |PS.n|/|n| = 9/sqrt(41) unit

my second attempt ::

the vector normal to the line, n1 = i+j-1/2k

the vector normal to the plane, n2 = i+2j+6k

n1.n2 = 0 ==> the two normal are perpendicular ===> the line and the plane are parallel

thus the distance for the line to the plane = sqrt(2^2+1^2+(-1/2)^2) = sqrt(21)/2 unit

WHICH ATTEMPT IS CORRECT SINCE I GOT TWO DIFFERENT ANSWERS??? PLEASE HELP ME..