# If the planes x=cy+bz , y=cx+az , z=bx+ay go through the straight line, then is it true that a^2+b^2+c^2+2abc=1?

Mar 9, 2017

Yes, it is true. Please see below for details.

#### Explanation:

Let the planes $x = c y + b z$ , $y = c x + a z$ , $z = b x + a y$ go through the straight line defined by $\left(p , q , r\right)$. The planes can also be written as

$x - c y - b z = 0$ , $c x - y + a z = 0$ , $b x + a y - z = 0$

As the planes pass through line $\left(p , q , r\right)$, the line is perpendicular to the normal of the plane, say $x - c y - b z = 0$ and hence dot product should be zero i.e.

$p - c q - b r = 0$

Similarly $c p - q + a r = 0$ and

$b p + a q - r = 0$

Solving them for $p , q$ and $r$ from first two equations, we get

$\frac{p}{- a c - b} = \frac{- q}{a + b c} = \frac{r}{- 1 + {c}^{2}}$

which implies $p = - a c - b$, $q = - a - b c$ and $r = {c}^{2} - 1$

and substituting in third we get

$b \left(- a c - b\right) + a \left(- a - b c\right) - \left({c}^{2} - 1\right) = 0$

or $- a b c - {b}^{2} - {a}^{2} - a b c - {c}^{2} + 1 = 0$

or ${a}^{2} + {b}^{2} + {c}^{2} + 2 a b c = 1$