Calculus: Early Transcendentals 8th Edition

$x=3t, y=1-t ,z=2-2t$
$L_1=\lt 0,1,2\gt+t\lt a,b,c\gt$; $x=at, y=1+bt, z=2+ct$ Here $v=\lt a,b,c\gt$ Normal of plane $n=\lt 1,1,1\gt$ Direction vector of line: $s=\lt 1,-1,2\gt$ Thus, $v-n=\lt1,1,1\gt$ and $v \cdot=\lt 1,-1,2\gt$ System of equations: $a+b+c=0$ and $a-b+2c=0$ Solve these equations to get the values for $a, b$ and $c$. $a=3,b=-1,c=-2$ Plug all these values in $x=at y=1+bt z=2+ct$, we get $x=3t, y=1-t ,z=2-2t$