Find Cartesian equations for $II_1$: $\left(\begin{matrix}x_1\\x_2\\x_3\\\end{matrix}\right)=\left(\begin{matrix}3\\-3\\-1\\\end{matrix}\right)+\mu_1\left(\begin{matrix}1\\-1\\1\\\end{matrix}\right)+\mu_2\left(\begin{matrix}2\\0\\-1\\\end{matrix}\right)$
My working out:
In order to find the cartesian equations for $\mu_1$ and $\mu_2$ , we first extract the equations from the plane, such that for $II_1$:
$x_1=3+\mu_1+2\mu_2\ldots\left(1\right)$
$x_2=-3\ -\ \mu_1\ldots\left(2\right)$
$x_3=-1+\mu_1\ -\ \mu_2\ldots\left(3\right).$
We then add equation $(1)$ and equation $(2)$ simultaneous equations:
$x_1=3+\mu_1+2\mu_2$
$+\ x_2=-3\ -\ \mu_1$
$x_1+x_2=2\mu_2$
$\mu_2=\frac{x_1}{2}+\frac{x_2}{2}\ldots\left(4\right).$
We can now rearrange equation $(3)$ to make $\mu_2\ $ the subject:
$\mu_2=-1+\mu_1\ -\ x_3.$
Hence, using equation $(4)$, we can deduce that:
$\frac{x_1}{2}+\frac{x_2}{2}=-1+\mu_1\ -\ x_3,$
which can then be rearranged to make the subject $\mu_1$:
$\mu_1=\frac{x_1}{2}+\frac{x_2}{2}+1+\ x_3...\ (5).$
Thus, substituting equation $(4)$ and equation $(5)$ into equation $(3)$ gives:
$x_3=-1+\left(\frac{x_1}{2}+\frac{x_2}{2}+1+\ x_3\right)-\left(\frac{x_1}{2}+\frac{x_2}{2}\right).$
The answer I got was just $0$... Can someone please explain where I went wrong?
Since you already used $(3)$ to get the value of $\mu_1$, substituting the parameters back in $(3)$ will give you a not so useful true statement, i.e. $0=0$.
Instead, you should use the unused equation $(2)$ :
$$\mu_1 =-3-x_2 =\frac{x_1}{2} +\frac{x_2}{2} +1+x_3 $$ Simplifying this will give you the cartesian equation.