How do you find distance from a point to a plane? Compute the distance from the point A = (1,1,-1) to the plane which contains the point B = (3,6,-1) and the line,  \begin{align}
    \begin{bmatrix}
           x \\
           y \\
           z \\
         \end{bmatrix} &= \begin{bmatrix}
           2 \\
           5 \\
           0 \\
         \end{bmatrix} t
  \end{align} 
NOT ALLOWED to use cross product to determine the answer!
The answer is 0, because point A is on the plane. But I don't know how to get the plane equation using the point B which is on the plane and the line which is also on the plane, to then derive the distance.
Edit: thanks for all the answers, I decided to go talk to my prof and she explained that we were supposed to first select two points on the line(lets call them G and H), so we would have three points on the plane, including point B. So we could then create two directional vectors, example: BG and BH. Using these two directional vectors, we would find a vector N, which is orthogonal to both of these vectors. this would give us a vector which is orthogonal to the plane. This would be done by have BG∙N=0 and BH∙N=0 and solving the system to give us N. After we get N, we will use N and a point on the plane, B to compute the distance from A to the plane.
 A: One way to write a general equation of a plane in $x,y,z$-space is
$$ ax + by + cz = d $$
where $a, b, c$, and $d$ are constants.
The equation is not unique, because you can multiply all the constants
by the same constant factor (as long as it is not zero) and get a different
equation of the same plane.
Three non-collinear points determine a plane in three-dimensional space.
From the problem description, you have at least these three points
on the plane:
\begin{align}
&(0,0,0) && \text{on the given line at $t=0$} \\
&(2,5,0) && \text{on the given line at $t=1$} \\
&(3,6,-1) && \text{the given point $B$}
\end{align}
Plugging each of these points into the general equation $ax + by + cz = d$
will give you a system of equations in which the only unknowns are
$a, b, c$, and $d$.
If the three points are not collinear, every solution of the system of equations will give you a set of constants $a, b, c$, and $d$
such that $ax + by + cz = d$ is a correct equation of the desired plane.
At this point you can just choose one of the solutions;
which one does not matter (at least not much)
as long as $a, b, c$, and $d$ are not all zero.
Note that this means the vector
$\begin{bmatrix} a & b & c \end{bmatrix}^T$
is perpendicular to the desired plane, and therefore it
is some scalar multiple of the cross product of any two independent
vectors parallel to the plane, which is the other way you could
have found an equation of the plane if you were allowed to use
a cross product.
In this particular case,
since the point $A$ actually is exactly in the plane (distance $0$),
just plugging the coordinates of $A$ into the equation will 
answer the question.
But in general, for any arbitrary point $(x,y,z)$,
the distance from $(x,y,z)$ to the plane is
$$
\frac{ax + by + cz - d}{\sqrt{a^2 + b^2 + c^2}}.
$$
A: Hint: The line contains the origin. 
Hence one direction vector is $u=(2,5,0)^T$ and the other $v=(3,6,-1)^T$, which is the vector from the origin to $B$.
The plane is given by $$x=\lambda u+\mu v.$$
$\lambda = -1$ and $\mu = 1$ gives the point $A$.
More general approach. calculate two different points on line and use them in combination with $B$ to set up the two directional vectors. This will only work if the line does not contain $A$, in which case the distance is 0 anyways and you cannot define a plane.
A: Any plane can be written in the form $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$ where $(x_0, y_0, z_0)$ is any point in the plane and  is any vector perpendicular to the plane. 
Here you are told that $(3, 6, -1)$ is a point in the plane so we must have $A(x- 3)+ B(y- 6)+ C(z+ 1)= 0$. Knowing that the line $x= 2t, y= 5t, z= 0$ is in the plane we know that $A(2t- 3)+ B(5t- 6)+ C= 0$ for all $t\in \mathbb{R}$. That can be written as $2At- 3A+ 5Bt- 6B+ C= (2A+ 5B)t+ (C- 3A- 6B)= 0$. Since that is to be true for all t, we must have $2A+ 5B= 0$ and $C- 3A- 6B= 0$. From $2A+ 5B= 0, B= (-2/5)A so C- 3A- 6B= C- 3A+ (12/5)A= C- (3/5)A= 0$. 
Of course, in the original form, $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$, we could divide through by any number, in particular $C$, and have an equation for the same plane.  So we can take $C= 1$ and have $1- (3/5)A= 0$ so $A= 5/3$.  Then $B= (-2/5)A= (-2/5)(5/3)= -2/3$.
The equation of the plane is $(5/3)(x- 3)- (2/3)(y- 6)+ (z+ 1)= 0$. Multiplying though by 3, $5(x- 3)- 2(y- 6)+ z+ 1= 5x- 15- 2y+ 12+ z+ 1= 0$ or $5x- 2y+ z= 2$.
A: You can take the vectors $[1,1,-1]^T$ and $[2,5,0]^T$ and see that they are spanning the plane ( why? ):
Now we try and solve
$$\left[\begin{array}{rr}1&2\\1&5\\-1&0\end{array}\begin{array}{|r}3\\6\\-1\end{array}\right]$$
We find a solution [1,1] which then are the coordinates of the point in the plane with respect to the vector.
This verification is a linear combination of the vectors spanning the plane: $$1\cdot \left[\begin{array}{c}1\\1\\-1\end{array}\right]
 + 1 \cdot \left[\begin{array}{c}2\\5\\0\end{array}\right]
 = \left[\begin{array}{c}1+2\\1+5\\-1+0\end{array}\right] = \left[\begin{array}{c}3\\6\\-1\end{array}\right]
$$
If the equation system was insolvable or had 0 solution, then we could have solved a least squares problem instead.
A: hint: Choose $2$ points on the plane: $t = 0 \implies A = (0,0,0)$, $t = 1\implies C = (2,5,0)$,and the given point $B = (3,6,-1)$, then find the normal line to the plane defined as $\vec{N} = \vec{AB} \times \vec{AC}= (m,n,p)$. You have to solve for $m,n,p$ as you do the cross product of the above two vectors. Then the formula for the plane is: $\vec{XB}\cdot \vec{N} = 0$, with $X = (x,y,z)$
