1. **Problem statement:** We have a line 𝓛 defined by parametric equations: $$x = 3 + 2t, \quad y = 2t, \quad z = t$$ and a point \(A(3,0,0)\). (a) Prove that \(A\) is on 𝓛. (b) Find two points \(E\) and \(F\) on 𝓛 each at distance 6 from \(A\). 2. **Step (a): Prove that A is on 𝓛** To check if \(A(3,0,0)\) lies on 𝓛, we need to find a parameter \(t\) such that: $$3 + 2t = 3, \quad 2t = 0, \quad t = 0$$ From \(2t=0\), we get \(t=0\). Substitute \(t=0\) into \(x\) and \(z\): $$x = 3 + 2(0) = 3, \quad z = 0$$ All coordinates match \(A\), so \(A\) is on 𝓛 at \(t=0\). 3. **Step (b): Find points E and F on 𝓛 at distance 6 from A** Let a point on 𝓛 be \(P(t) = (3+2t, 2t, t)\). Distance from \(A(3,0,0)\) to \(P(t)\) is: $$d = \sqrt{(3+2t - 3)^2 + (2t - 0)^2 + (t - 0)^2} = \sqrt{(2t)^2 + (2t)^2 + t^2}$$ Simplify inside the square root: $$= \sqrt{4t^2 + 4t^2 + t^2} = \sqrt{9t^2} = 3|t|$$ Set distance \(d=6\): $$3|t| = 6 \implies |t| = 2$$ So \(t = 2\) or \(t = -2\). Calculate points: - For \(t=2\): $$E = (3 + 2(2), 2(2), 2) = (7, 4, 2)$$ - For \(t=-2\): $$F = (3 + 2(-2), 2(-2), -2) = (-1, -4, -2)$$ **Final answers:** - (a) \(A(3,0,0)\) is on 𝓛 at \(t=0\). - (b) Points \(E(7,4,2)\) and \(F(-1,-4,-2)\) lie on 𝓛 and are each 6 units from \(A\).