Question

What is the nature of the intersection of the set of planes $x+ay+(b+c)z+d=0,x+by+(c+a)z+d=0$ and $x+cy+(a+b)z+d=0$?

• A. They meet at a point.
• B. They form a triangular prism.
• C. They pass through a line.
• D. They are at equal distance from the origin.

Answer 1

Answer: (C)

They pass through a line.

Given planes are
$x+ay+(b+c)z+d=0,x+by+(c+a)z+d=0$ and $z+cy+(a+b)z+d=0$
The rectangular array (or matrix) is
$\left[\begin{array}{cccc}1& a& b+c& d\\ 1& b& c+a& d\\ 1& c& a+b& d\end{array}\right]$
Since all the minors of third order of matrix are zero, in other words all square sub-matrices of order $3$ are singular.
Therefore planes intersect in a line i.e pass through a line.