Prove that the plane through the point $(\alpha, \beta, \gamma)$ and the line $x = py + q = rz + s$ Prove that the plane through the point $(\alpha, \beta, \gamma)$ and the line $x = py + q = rz + s$ is
$${\begin{vmatrix} x& py+q& rz+s\\
\alpha& p\beta+q& r\gamma+s\\
1& 1& 1\\
\end{vmatrix}} = 0$$
My Attempt:
The equation of line is
$x = py + q = rz + s$
$$\implies x = p (y + \frac {q}{p}) = r ( z + \frac {s}{r}) $$
$$\implies \frac { x - 0}{1} = \frac {y + \frac {q}{p}}{\frac {1}{p}} = \frac {z + \frac {s}{r}}{\frac {1}{r}}$$
The equation of plane through this line is
$$a(x-0) + b(y+\frac {q}{p}) + c(z+\frac {s}{r}) = 0$$
where $a,b,c$ are the direction ratios of the line normal to the plane.
Now,
$$a\cdot 1 + b\cdot (y+\frac {q}{p}) + c\cdot (z + \frac {s}{r}) = 0$$
Also, the plane passes through the point $(\alpha, \beta, \gamma)$,
$$a(\alpha - 0) + b(\beta + \frac {q}{p}) + c(\gamma + \frac {s}{r}) = 0$$
How to proceed further from here?
 A: SAs per OP's work the line passes throught $A(0,-q/p,-s/r)$ its direction vector is $\vec L=(1,1/p,1/r)$. The plane passes through $B(u,v,w)$. Normal to the plane $\vec n=\vec{AB}\times \vec L=(n_1,n_2,n_3)$, then the equation of the plane is
$$n_1(x-u)+n_2(y-v)+n_3(z-w)=0.$$ This equation can also be written as
$$(\vec r-\vec B).(\vec{AB} \times \vec L)=0 \implies\begin{vmatrix} x-u & y-v & z -w  \\ 0-u & -q/p-v & -s/r-w \\ 1 & 1/p & 1/r \end{vmatrix}=0,$$ that gives the equation of the required plane.
Next, $C_2'\to p C_2, C_3' \to rC_3$ to get
$$\begin{vmatrix} x-u & py-pv & rx-rw \\ u & q+vp & s+rw \\ 1 & 1 & 1\end{vmatrix}=0$$
Next, Perform $R_1'\to R_1+R_2$ to get the required form:
$$\begin{vmatrix} x & py+q & rx+s \\ u & vp+q & rw+s \\ 1 & 1 & 1\end{vmatrix}=0$$
A: Hint:
$\frac { x - 0}{1} = \frac {y + \frac {q}{p}}{\frac {1}{p}} = \frac {z + \frac {s}{r}}{\frac {1}{r}}$
One of the points on the line is $P \,$ $(0,-\frac{q}{p}, -\frac{s}{r}$).
Direction vector from this point to $Q \,$ $(\alpha, \beta, \gamma)$ is $\vec{PQ} \, (\alpha, \beta + \frac{q}{p}, \gamma + \frac{s}{r})$.
As the given line and $\vec{PQ}$ are in the plane we take a cross product of vectors $\, (\alpha, \beta + \frac{q}{p}, \gamma + \frac{s}{r})$ and $(1, \frac{1}{p}, \frac{1}{r}) \,$to find the normal vector to the plane say, $(a, b, c)$.
Then the equation of the plane is $a(x-\alpha) + b(y-\beta) + c(z-\gamma) = 0$
