Question

If the planes $x-bz=0,cx-y+=0$ and $bx+ay-z=0$, pass through a line, then find the value of $a^2+b^2+c^2+2abc$.

• A. 0
• B. 1
• C. $-1$
• D. $\frac{1}{2}$

Answer 1

The correct option is B 1

Given, planes are
$x-cy-bz=0$...(i)
$a-y+az=0$ ...(ii)
$bx+ay-z=0$ ...(iii)
Equation of plane passing through the line of intersection of panes (i) and (ii) maybe taken as
$(X-cy-bz)+\lambda (cx-y+az)=0$
$(1+c\lambda )x+y(-c-\lambda )+z(-b+a\lambda )=0$ ...(iv)
Now, planes (iii) and (iv) are same.
$\therefore \frac{1_c\lambda }{b}=\frac{-(c+\lambda )}{a}=\frac{-b+a\lambda }{-1}$
By elimintaing $\lambda$, we get
$a^2+b^2+c^2+2abc=1$