Find the value(s) of $k$ so that the following set of planes has a line of intersection. Question help?

Question

Here are the equations:

• $2x + 7y + 2z = 3$,
• $6x + y - kz = -1$,
• $2x + (2k + 1)y + (k - 1)z = 4$.

Help on this question would be greatly appreciated!

Answer 1

First of all, we need the determinant $$\left|\begin{matrix} 2 & 7& 2\\ 6&1&-k\\ 2&2k+1&k-1 \end{matrix} \right|=0 $$ or $$2(k-1)-14k+12(2k+1)-2+2k(2k+1)-42(k-1)=0 $$ from this we have $$4k^2-24k+36=0 $$ which has solution $k=3$.

Now replace $k=3$ to the equation of planes to see if this planes intersect in a line or not.