Step 1

Given, the two planes x – y – z = 1 and 2x + 3y + z = 2, we have to find the parametric equations for the intersections of these two planes.

Step 2

To find a point on this line we can for instance set z = 0 and then use the given equations to solve for x and y.

x-y=1

2x+3y=2

On solving these two equations we get x = 1, y = 0. So the point on the line is (1, 0, 0).

Also, the direction of the line lives in both the planes and so, in particular, is perpendicular to both normal vectors, therefore a vector which is parallel to the line is given by

\((1,-1,-1)\times(2,3,1)=\begin{vmatrix}i & j & k\\1 & -1 & -1 \\ 2 & 3 & 1\end{vmatrix}=i(-1+3)-j(1+2)+k(3+2)\)

=2i-3j+5k

=(2,-3,5)

Thus the equation of the line is given by the vector equation

(x,y,z)=(1,00)+t(2,-3,5)

=(1+2t,-3t, 5t)

Hence the parametric equations are x = 1 + 2t, y = -3t, and z = 5t.