There is a problem in my book
In the question the circles are given in the form $S+kP=0$ where $S$ is plane and $P$ is a plane and $k$ is real number. The question asks us to prove that the circles lie on the same sphere and to find the equation of the sphere .
I can answer only the second part. I asked my teacher and he said that we can't prove the first part. Any thoughts or suggestions?? Thanks.
Let us type all data here.
$$ \begin{aligned} S_1(x,y,z) &= x^2 +y^2+z^2 - 2x+3y+4z-5\ ,\\ P_1(x,y,z) &= 5y+6z+1\ ,\\[2mm] S_2(x,y,z) &= x^2 +y^2+z^2 - 3x-4y+5z-6\ ,\\ P_2(x,y,z) &= x+2y-7z\ ,\\[2mm] (C_1)&=\{\ (x,y,z)\ :\ S_1(x,y,z)=P_1(x,y,z)=0\ \}\ ,\\ (C_2)&=\{\ (x,y,z)\ :\ S_2(x,y,z)=P_2(x,y,z)=0\ \}\ . \end{aligned} $$ The following solution supports the geometric intuition.
$S_1=0$ describes a sphere centered in $\Omega_1=\displaystyle \left(1,-\frac 32,-2\right)$ with radius $\frac 72$. We cut it with a plane, $P_1=0$, getting a circle. If this circle is on some other sphere, then this other sphere is centered in a point $\Omega$ on the normal form $\Omega_1$ to the plane. We already see the plane and its normal, so $\Omega$ is on the line with points parametrized as $\displaystyle \left(1,-\frac 32,-2\right)+\lambda(0,5,6)$, $\lambda\in\Bbb R$.
The same game for the other circle. $S_2=0$ describes a sphere centered in $\Omega_2=\displaystyle \left(\frac 32,2, -\frac52\right)$ with radius $4$. We cut it with a second plane, $P_2=0$, getting a circle. If this circle is on the same sphere above, then $\Omega$ is on the normal form $\Omega_2$ to the plane. So $\Omega$ is on the line with points parametrized as $\displaystyle \left(\frac 32,2,-\frac52\right)+\mu(1,2,-7)$, $\mu\in\Bbb R$.
Do these two lines intersect? Yes, $\lambda=\frac 12$, $\mu=-\frac 12$, lead to $\Omega=(1,1,1)$. So far we land so far as using the hint. ($\lambda,\mu$ are up to a factor two some $\lambda_1,\lambda_2$ special values from the hint.)
We still have to check the "radius match". Here we should go back to algebraic geometry.
A point is on $(C_1)$ if it is satisfying for all $\lambda_1$ the equation $S_1+\lambda_1 P_1=0$.
A point is on $(C_2)$ if it is satisfying for all $\lambda_2$ the equation $S_2+\lambda_2 P_2=0$.
Now use $\lambda_1=2\lambda$, $\lambda_2=2\mu$ (some "equation dedoubling" is the reason for short) and check that we have the same equation. (This corresponds to a "radius match", algebraically to a match of the constant coefficient.)