Answer: (a) \(5x-2y+z=-14\)

(b) \(\dfrac7{\sqrt{30}}\) \((\approx1.28)\)

(c) \(\mathbf r=\begin{pmatrix}0\\7\\0\end{pmatrix}+\lambda\begin{pmatrix}0\\1\\2\end{pmatrix}\)

(a)

The direction vectors in the plane are

\(\mathbf a=(0,1,2),\qquad \mathbf b=(1,3,1).\)

A normal vector is

\(\mathbf a\times\mathbf b=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&1&2\\1&3&1\end{vmatrix}=(-5,2,-1).\)

Use the equivalent normal vector \((5,-2,1)\). The point \((-3,-1,-1)\) lies on the plane. Hence

\(5(x+3)-2(y+1)+(z+1)=0.\)

So

\(5x-2y+z=-14.\)

(b)

The point is \((-1,0,-2)\), and the plane is \(5x-2y+z+14=0\). Therefore

\(d=\frac{|5(-1)-2(0)+(-2)+14|}{\sqrt{5^2+(-2)^2+1^2}}=\frac7{\sqrt{30}}.\)

(c)

A common point is found by taking \(x=0\), \(z=0\). Then \(3x+2y-z=14\) gives \(y=7\), so \((0,7,0)\) lies on both planes.

The direction vector of the intersection line is perpendicular to both normal vectors \((5,-2,1)\) and \((3,2,-1)\):

\(\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\5&-2&1\\3&2&-1\end{vmatrix}=(0,8,16)\sim(0,1,2).\)

Hence

\(\mathbf r=\begin{pmatrix}0\\7\\0\end{pmatrix}+\lambda\begin{pmatrix}0\\1\\2\end{pmatrix}.\)

Mark scheme