
Planar intersection
so I have a question I cannot answer in my calculus/vectors course, and it's driving me insane. if someone could just tell me how to find k i would be forever grateful.
'Find the value of k such that the three planes will always intersect in a point, then find the point of intersection.'
π1: x+2yz5=0
π2: x+ky3z+11=0
π3 2x+y+z+10=0

From a linear algebra perspective, you get the intersection of a point if the system of equations you have there has exactly one solution, right? And when does that happen for the system
$\displaystyle A\mathbf{x}=\mathbf{b},$ where
$\displaystyle A=\begin{bmatrix}1 &2 &1\\ 1 &k &3\\ 2 &1 &1\end{bmatrix}\qquad\text{and}\qquad\mathbf{b}=\b egin{bmatrix}5\\11\\10\end{bmatrix}?$

Thats my problem, i'm drawing a complete blank. I've tried all the methods I know to narrow it down, and so far all I know is k cannot be 4

I wouldn't call that a complete blank! Especially since it's correct. k = 4 makes the matrix A singular, which means you'd get at least a line of intersection, not a single point.
What's puzzling to me, and it sounds like to you, is that the problem asks for "the" value of k that makes the intersection just a point. That sounds to me like a typo. There are infinitely many values of k that make the intersection one unique point. Just with my own investigations here, solving the system by leaving k unspecified yields different solutions that depend on k.
That's the best I can do for you, I'm afraid. I think it's a typo. They should have asked "What value of k makes the intersection a line?"