Find the equation of the plane passing through the point $\left(1, -1, -1\right)$ and perpendicular to each of the planes $2 \left(x - 2y\right) + 2 \left(y - z\right) - \left(6 z + x\right) = 0$ and $2 \left(y - x\right)+ 3 \left(y + z\right) + 4 \left(x - z\right) = 0$
The given planes are:
$2 \left(x - 2y\right) + 2 \left(y - z\right) - \left(6 z + x\right) = 0$
i.e. $\;$ $2 x - 4y + 2y - 2z - 6z - x = 0$
i.e. $\;$ $x - 2y - 8z = 0$ $\;\;\; \cdots \; (1a)$
and $\;$ $2 \left(y - x\right)+ 3 \left(y + z\right) + 4 \left(x - z\right) = 0$
i.e. $\;$ $2y - 2x + 3y + 3z + 4x - 4z = 0$
i.e. $\;$ $2x + 5y - z = 0$ $\;\;\; \cdots \; (1b)$
The required plane passes through the point $\left(1, -1, -1\right)$
$\therefore \;$ Let the equation of the required plane be
$a \left(x - 1\right) + b \left(y + 1\right) + c \left(z + 1\right) = 0$ $\;\;\; \cdots \; (2)$
Since plane $(2)$ is perpendicular to each of the planes $(1a)$ and $(1b)$, we have,
$a - 2b -8c = 0$ $\;\;\; \cdots \; (3a)$ and
$2a + 5b -c = 0$ $\;\;\; \cdots \; (3b)$
[Note: for two perpendicular planes $a_1 x + b_1 y + c_1 z + d_1 = 0$ and $a_2 x + b_2 y + c_2 z + d_2 = 0$, we have $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$]
Solving equations $(3a)$ and $(3b)$ we get,
$\dfrac{a}{\left(-2\right) \left(-1\right) - \left(5\right) \left(-8\right)} = \dfrac{b}{\left(1\right) \left(-1\right) - \left(2\right) \left(-8\right)} = \dfrac{c}{\left(1\right) \left(5\right) - \left(2\right) \left(-2\right)} = \lambda$ (say)
i.e. $\;$ $\dfrac{a}{42} = \dfrac{b}{15} = \dfrac{c}{9} = \lambda$
i.e. $\;$ $a = 42 \lambda$, $\;$ $b = 15 \lambda$, $\;$ $c = 9 \lambda$
Substituting the value of $a$, $b$ and $c$ in equation $(2)$, the equation of the required plane is
$42 \lambda \left(x - 1\right) + 15 \lambda \left(y + 1\right) + 9 \lambda \left(z + 1\right) = 0$
i.e. $\;$ $42 x - 42 + 15 y + 15 + 9z + 9 = 0$
i.e. $\;$ $42x + 15y + 9z = 18$
i.e. $\;$ $14x + 5y + 3z = 6$