Question

The planes $x = cy + bz,y = az + cx,z = bx + ay$ passes through one line , if
A. $a + b + c = 0$
B. $a + b + c = 1$
C. ${a^2} + {b^2} + {c^2} = 1$
D. ${a^2} + {b^2} + {c^2} + 2abc = 1$

Answer 1

Hint: We will let plane ${P_1}$ as $x = cy + bz$, plane ${P_2}$ be $y = az + cx$ and ${P_3}$ be $z = bx + ay$. Then, since all the plane passes through one line, the determinant of the coefficients of the equations of the plane must be 0. Write the determinant and form the equation by expanding the determinant along the first row.

Complete step-by-step answer:
We are given 3 planes, $x = cy + bz,y = az + cx,z = bx + ay$
Let plane ${P_1}$ be $x = cy + bz$ which is also equal to $x - cy - bz = 0$
Let plane ${P_2}$ be $y = az + cx$ which is also equal to $cx - y + az = 0$
And let plane ${P_3}$ be $z = bx + ay$ which is also equal to \[bx + ay - z = 0\]
Now, we are given that all these planes pass through one line.
Then, the determinant of the coefficients of the equations of the plane should be 0.
$\left| {\begin{array}{*{20}{c}}
  1&{ - c}&{ - b} \\
  c&{ - 1}&a \\
  b&a&{ - 1}
\end{array}} \right| = 0$
We will expand the determinant through the first row.
$
  1\left( {1 - {a^2}} \right) + c\left( { - c - ab} \right) - b\left( {ac + b} \right) = 0 \\
   \Rightarrow 1 - {a^2} - {c^2} - abc - abc - {b^2} = 0 \\
   \Rightarrow {a^2} + {b^2} + {c^2} = 1 - 2abc \\
$
On rearranging it further, we will get,
\[{a^2} + {b^2} + {c^2} + 2abc = 1\]
Hence, option D is the correct answer.

Note: Note: While writing the entries in determinant, be careful about the coefficients. The coefficient of $x$ should be one column, coefficient of $y$ in one column and coefficient of $z$ in last column. We have opened the determinant through the first row, but it can be expanded through any row or any column.