Write the equation of a plane passing through points $A(1, 0, 1)$ and $B(0, -1, 1)$ and located at a distance of $4$ from point $C(5, 0, -3)$.
What steps I should take to solve this question as after putting value of the length I am not getting anything from that.
I have tried solving by taking one point M on the plane, such that MC becomes perpendicular to the plane. And using it as normal to write the equation of the plane with point and normal equation form… but I didn't work as there remains many variables.
Best Answer
Alternative solution:
Let $\pi \ldots ax+by+cz+d=0$ is the required plane.
$A \in \pi \implies a + c + d = 0 \quad\qquad (1)$
$B \in \pi \implies -b+c+d=0 \,\, \qquad (2)$
$(1)-(2) \iff a + b = 0 \iff \boxed{a = -b}$.
Next, we will use now an another condition, i.e. $d(C, \pi) = 4$, or $$ \left| \frac{5a-3c+d}{\sqrt{a^2+b^2+c^2}} \right| = 4, $$ or $$ \left| \frac{4a-4c+\overbrace{a+c+d}^0}{\sqrt{a^2+a^2+c^2}} \right| = 4, $$ or $$ 4|a-c| =4\sqrt{2a^2+c^2}, $$ or $$ 16a^2+32ac=0 \iff 16a(a+2c)=0 \iff \boxed{a=0} \text{ or } \boxed{a=-2c}. $$ Finally, in the first case, i.e. for $a=0$, we have $b=0$, and form $(1)$ we can get $c=-d$.
In the second case, i.e. for $a=-2c$ from $(1)$ we can get $c=d$, and then from $(2)$ we will have $b=2d \, (=2c=-a)$.
Hence, there are two required planes:
$\pi_1 \ldots z - 1 = 0$,
$\pi_2 \ldots 2x-2y-z-1=0$.