Question

Assertion :The three planes $x+ay+(b+c)z+d=0$, $x+by+(a+c)z+d=0$, $x+cy+(a+b)z+d=0$ have a common line. Reason: The point $(-d,0,0)$ is common to all the three planes.

• A. Both Assertion & Reason are individually true & Reason is correct explanation of Assertion
• B. Both Assertion & Reason are individually true but Reason is not the correct (proper) explanation of Assertion
• C. Assertion is true but Reason is false
• D. Assertion is false but Reason is true

Answer 1

The correct option is A Both Assertion & Reason are individually true & Reason is correct explanation of Assertion

For the point of intersection of planes
$x+ay+(b+c)z+d=0,x+by+(c+a)z+d=0$

Putting $z=0$, we get $x+ay+d=0$ & $x+by+d=0$
Solving these two equations, we get $x=-d,y=0$
$\therefore$ point on the above two planes is $(-d,0,0)$, which lies on the plane $x+cy+(a+b)z+d=0$, so point $(-d,0,0)$ is common to all the three planes.
$\Rightarrow$ Reason(R) is true
Again, let $l,m,n$ be DC's of common line.
$\therefore$ $l+ma+n(b+c)=0$ and $l+mb+n(a+c)=0$
$\Rightarrow \frac{l}{a(a+c)-b(b+c)}=\frac{-m}{a+c-b-c}=\frac{n}{-(a-b)}$
$\Rightarrow \frac{l}{a(a+c)-b(b+c)}=\frac{-m}{-(a-b)}=\frac{n}{-(a-b)}$
$\Rightarrow \frac{1}{a+b+c}=\frac{m}{-1}=\frac{n}{-1}$
Now equations of common line is $\frac{x+d}{a+b+c}=\frac{y-0}{-1}=\frac{z-0}{-1}$
and normal to the plane $x+cy+(a+b)z+d=0$ is $\perp$ to common line.