Find the equation of a plane that contains a given line and is 2 units away from a given point. The given line has a vector of (2,-1,4) and contains the points A(-1,1,-1) and B(1,0,3). The given  point is P(2,3,4). The problem asks you to find a plane such that its distance to P is 2 units and it contains line AB, and to justify the number of solutions it has.
I've tried making the plane's orthogonal vector (A,B,C) unitary and multiplying it by 2 to make its modulus 2, but I don't know how to find A, B and C from there.
Another approach is using the distance formula, but I get a single equation with 4 variables, some of which are squared. 
 A: First find all the planes through the given line as follows.(We speak of the "pencil" of planes through the given line.)Pick an arbirary point $(x_1,y_1,z_1)$ and find the equation of the plane through $(-1,1,-1),(1,0,3) \text { and } (x_1,y_1,z_1)$ in the form $$ax+by+cz+d=0.$$ N.B. Remember that if you find an equation of a plane as 
$$\alpha x+\beta y+\gamma z+\delta=0$$ then, for any $k \ne 0,$ $$k\alpha x+k\beta y+k\gamma z+k\delta=0$$ is the same plane, so you can get rid of  awkward fractions and force $a,b,c,d$ to be integers. Now pick another arbirary point $(x_2,y_2,z_2)$ and find the equation of the plane through $(-1,1,-1),(1,0,3) \text { and } (x_2,y_2,z_2)$ in the form $$a'x+b'y+c'z+d'=0.$$ Then a plane through the line $\ell$ has the form $$(a+\lambda a')x+(b+\lambda b')y+(c+\lambda c')z+(d+\lambda d')=0,$$ or is the plane $$a'x+b'y+c'z+d'=0$$ which has to be treated separately as a special case. The distance from (2,3,4) to the plane $$(a+\lambda a')x+(b+\lambda b')y+(c+\lambda c')z+(d+\lambda d')=0$$ is $$D=\frac{|(a+\lambda a')2+(b+\lambda b')3+(c+\lambda c')4+d+\lambda d'|}{\sqrt{(a+\lambda a')^2+(b+\lambda b')^2+(c+\lambda c')^2}}$$ The only unknown in $D$ is $\lambda.$Yo need $D$ to be 2, so solve $D^2=4$, which is a quadratic equation in $\lambda$ and put this value of $\lambda$ back into $$(a+\lambda a')x+(b+\lambda b')y+(c+\lambda c')z+(d+\lambda d')=0$$ as your answer. To be really thorough, you should also find the distance from (2,3,4) to the plane $$a'x+b'y+c'z+d'=0$$ just in case, by some fluke, that distance happens to be 2.
A: Find two unit vectors $u$ and $v$ that are perpendicular to each other and to $(2,-1,4)$.  The plane's unit normal is $w=u\cos\theta+v\sin\theta$.
The distance is a dot product; solve for theta.
A: Let $T$ be a point of the sphere $s$ centered at $P$ withe radius $2$ such that the distance from the plane $p$ defined by $T$, $A$ and $B$ to $P$ is equal to $2$. Then $p$ is tangent to $s$ and therefore the vectors $TA$ and $TB$ are orthogonal to $TP$. So, you can find the point $T$ by solving the system$$\left\{\begin{array}{l}(x-2)^2+(y-3)^2+(z-4)^2=4\\(x-2)(x+1)+(y-3)(y-1)+(z-4)(z+1)=0\\(x-2)(x-1)+(y-3)y+(z-4)(z-3)=0,\end{array}\right.$$which is equivalent to$$\left\{\begin{array}{l}x^2-4 x+y^2-6 y+z^2-8 z+25=0\\x^2-x+y^2-4 y+z^2-3 z-3=0\\x^2-3 x+y^2-3 y+z^2-7 z+14=0.\end{array}\right.$$Now, if you subtract the first equation from each of the other two, you get the system$$\left\{\begin{array}{l}x^2-4 x+y^2-6 y+z^2-8 z+25=0\\3 x+2 y+5 z=28\\x+3 y+z=11.\end{array}\right.$$Solving the last two equations with respect to $x$ and $y$, you get that$$\left\{\begin{array}{l}x=\frac{62-13z}7\\y=\frac{5+2z}7.\end{array}\right.$$Finally, you put these values of $x$ and $y$ in the first equation in order to get a quadratic equation in $z$, whose roots are$$z=\frac{426\pm7\sqrt{138}}{111}.$$So, the solutions are$$\frac1{111}\left(192\mp13\sqrt{138},201\pm2\sqrt{138},426\pm7\sqrt{138}\right).$$
