Find a plane passing through a point and parallel to 2 different vectors.

Question

Ok so this is going to be on exam,and I have tried a multitude of methods of solving it.Namely I tried mostly by solving systems obtained by some weird substitions.

Here is how it goes.Find an equation of a plane such that it passes through point $A(3,4,-5)$ and it is parallel to vectors $a=\langle 3,1,-1 \rangle$ and $b=\langle 1,-2,1 \rangle$.

I have tried by substituting coordinates of the point A into general form of plane equation,and by noting that if $n = \langle A ,B,C \rangle$ is vector orthogonal to our needed plane then it is also orthogonal to both vectors a and b,thus their dot product must be zero.Here is what I get:

$$3A + 4B -5C =0$$ $$A -2B +C =0$$ $$3A+B-1C = 0$$

But solving this system yields that all coefficients are zero and that such plane does not exist,but it does since our teacher has given us the sollution.

Answer 1

Found an answer.Basically what we need is a vector that is orthogonal to both vectors $a$ and $b$,we can obtain such a vector and it is equal to $a \times b $.Now we use the obtained vector as an vector orthogonal to desired plane,and thus we find the necessary plane

Answer 2

Alternative solution:

The equation of the plane $\Pi_1$ which is defined by the 2 linearly independent vectors $\vec a= (3,1,-1)$ and $\vec b = (1,-2,1)$ is:

$$\begin{vmatrix}x & y & z \\ 3 & 1 & -1 \\ 1 &- 2 & 1 \end{vmatrix}=0\implies-x-4y-7z=0$$

So, we have $\Pi_1 :-x-4y-7z=0$

Now, we know that if there is $d'\in \mathbb R$, such that $\Pi_1:ax+by+cz=d$ and $\Pi_2:ax+by+cz=d' \iff \Pi_1 \parallel \Pi_2$.

So, $\Pi_2: -x-4y-7z=d'$. We already know that $(3,4,-5) \in \Pi_2\implies d'=16$.

Done.