Solution:
\( \displaylines{\text{Normal is }<2,3,6>\text{ which is a vector perpendicular to the plane.}\\ <2t,3t,6t>+<0,0,0>\text{ is a path which contains the origin and travels perpendicular to the plane.}\\ \text{Solve for }t\text{ when this path intersects the plane:}\\ 2\left(2t\right)+3\left(3t\right)+6\left(6t\right)-10=0\\ 4t+9t+36t-10=49t-10=0\implies t=\frac{10}{49}\\ x=2t=\frac{20}{49}\\ y=3t=\frac{30}{49}\\ z=6t=\frac{60}{49}} \)