2025 ACJC P1 Q9

The planes \(\pi_1\) and \(\pi_2\) have equations \(3x+c(y+z)-2=0\) and \(\mathbf{r}=(\mathbf{i}+3\mathbf{j}-2\mathbf{k})+s(2\mathbf{i}-\mathbf{j}+3\mathbf{k})+t(\mathbf{i}-\mathbf{k})\) respectively, where \(c\) is a constant, and \(s\) and \(t\) are parameters. The point \(A(1,3,-2)\) lies in both planes.

1. Show that \(c=-1\).[1]
2. Show that the vector equation of the line of intersection of \(\pi_1\) and \(\pi_2\), line \(l\), is given by \(\mathbf{r}=\mathbf{i}+3\mathbf{j}-2\mathbf{k}+\alpha(\mathbf{i}-\mathbf{j}+4\mathbf{k})\), where \(\alpha\) is a parameter.[3]
3. Find the position vectors of the points on the line \(l\) which are a distance of \(3\sqrt{2}\) from the point \(B(2,-3,7)\).[4]
4. Find the equation of the plane \(\pi_3\) which is parallel to \(\pi_2\) and contains the point \(B\).
   Hence show that the distance between the planes \(\pi_2\) and \(\pi_3\) is \(\frac{20}{3\sqrt{3}}\).[3]