Classify a line relative to a plane before solving

With normal $\mathbf n=(2,1,-3)$ and direction $\mathbf d=(-2,1,-3)$, $\mathbf n\cdot\mathbf d=-4+1+9=6\ne0$, so there is a unique intersection. Substitution gives $6-4\lambda-1+\lambda-3+9\lambda=5$, hence $6\lambda=3$, $\lambda=1/2$. The point is $(2,-1/2,-1/2)$.

Both normal and direction have magnitude $\sqrt{14}$. Their angle is $\cos^{-1}(6/14)=1.1279$ rad (5 s.f.), or $64.623^\circ$ (3 d.p.). The line–plane angle is $\pi/2-\cos^{-1}(3/7)=\sin^{-1}(3/7)\approx0.443$ rad (3 s.f.), or $25.4^\circ$ (1 d.p.).

$(2,1,-3)\cdot(1,1,1)=0$, so the line direction is parallel to the plane. Its base point satisfies $(2,1,0)\cdot(2,1,-3)=5$. Therefore every point of the line lies on the plane; there is no unique intersection.