Plane equation using 3 points

Since the points $P$ , $Q$ , and $R$ on the plane then

$\begin{cases} P: & a + 3b + 6c = 7 & ...(1) \\ Q: & 2a + 5b + 7c = 7 & ...(2) \\ R: & -a + 4b - 6c = 7 & ...(3) \end{cases}$

From $(1) + (3): \ 7b = 14 \implies b = 2$ .

From $(2)-(1)+(3): \ 6(2) - 5c = 7 \implies - 5c = - 5 \implies c = 1$

From $(1): \ a + 3(2) + 6(1) = 7 \implies a = - 5$ .

Therefore $a+b+c = -5 + 2 + 1 = \boxed {-2}$ .

the given points satisfy the equation of the given plane. The goal is to find $a$ , $b$ and $c$ . Using this information, one gets:

$a + 3b + 6c = 7$ $2a + 5b + 7c = 7$ $-a + 4b - 6c = 7$

This can be written as:

$\left[\begin{matrix}1&3&6\\2&5&7\\-1&4&-6\end{matrix}\right]\left[\begin{matrix}a\\b\\c\end{matrix}\right]=\left[\begin{matrix}7\\7\\7\end{matrix}\right]$

$\implies \left[\begin{matrix}a\\b\\c\end{matrix}\right] = \left[\begin{matrix}1&3&6\\2&5&7\\-1&4&-6\end{matrix}\right]^{-1}\left[\begin{matrix}7\\7\\7\end{matrix}\right]$

The quantity $a+b+c$ can be found by performing the following trick:

$a+b+c = \left[\begin{matrix}1&1&1\end{matrix}\right]\left[\begin{matrix}a\\b\\c\end{matrix}\right]$ $a+b+c = \left[\begin{matrix}1&1&1\end{matrix}\right]\left[\begin{matrix}1&3&6\\2&5&7\\-1&4&-6\end{matrix}\right]^{-1}\left[\begin{matrix}7\\7\\7\end{matrix}\right]$ $\boxed{\implies a+b+c = -2}$

Ofcourse, there are more standard approaches to the problem like using vectors or simply solving the equation set without linear algebra. Calculations and simplifications have been left out in this solution.

Let the equation of the plane be $ax+by+cz=d$ . Since the points $(1,3,6),(2,5,7), (-1,4,-6)$ lie on the plane, we have

$a+3b+6c=d$

$2a+5b+7c=d$

$-a+4b-6c=d$

Adding first and third equations we get $7b=2d\implies b=\dfrac{2d}{7}$

Substituting in the first equation we get $a+6c=d-\dfrac{6d}{7}=\dfrac{d}{7}$ . Substituting in the third equation we get $2a+7c=d-\dfrac{10d}{7}=-\dfrac{3d}{7}$ .

From these two we get $c=\dfrac{d}{7}$ and $a=-\dfrac{5d}{7}$ .

So the equation of the plane is $-\dfrac{5d}{7}x+\dfrac{2d}{7}y+\dfrac{d}{7}z=d$ or $-5dx+2dy+dz=7d$ . In the simplest form, this equation reads as $-5x+2y+z=7$ , which is given in the problem. So $a=-5, b=2, c=1$ and $a+b+c=-5+2+1=\boxed {-2}$