Determine λ so that the planes x-y+z=0 , 3x-y-z+2=0…

Determine $\lambda$ so that the planes $x-y+z=0$, $3x-y-z+2=0$, $4x-y-2z+\lambda=0$ intersect along the same line.

Step 1: Coefficient matrix

\[ \Delta=\begin{vmatrix}1&-1&1\\3&-1&-1\\4&-1&-2\end{vmatrix}=1-2+1=0, \]

\[ \begin{vmatrix}1&-1\\3&-1\end{vmatrix}=2\ne0 . \]

So $\operatorname{rank}M=2$ for every $\lambda$.

Step 2: Consistency

Write $(4,-1,-2)=\mu(1,-1,1)+\nu(3,-1,-1)$. This gives $\mu+3\nu=4$, $\mu+\nu=1$ and $\mu-\nu=-2$, so $\mu=-\tfrac12$ and $\nu=\tfrac32$. The free terms must satisfy the same relation:

\[ \lambda=\mu\cdot0+\nu\cdot2=3 . \]

\[\lambda=3\]

Step 3: Common line

\[x=t-1,\quad y=2t-1,\quad z=t\qquad\big(\vec d=\{1,2,1\}\big)\]

Check: $\vec n_1\times\vec n_2=\{1,-1,1\}\times\{3,-1,-1\}=\{2,4,2\}\parallel\{1,2,1\}$.

The point $(-1,-1,0)$ satisfies all three equations, e.g. $-4+1+3=0$.