Question

If the planes $x–cy–bz=0$, $cx–y+az=0$ and $bx+ay–z=0$ pass through a straight line, then$a^2+b^2+c^2+2abc$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B

$1$

Explanation for the correct answer:

The equations of the given planes are

$P1:$$x–cy–bz=0$ $...\left(i\right)$

$P2:$$cx–y+az=0$ $...\left(ii\right)$

$P3:$$bx+ay–z=0$ $...\left(iii\right)$

Equation of planes passing through the line of intersection of planes $P1$ and $P2$ is given as

$x–cy–bz+$$\lambda \left(cx–y+az\right)=0$

Rearranging the terms we get

$\Rightarrow$$x\left(1+c\lambda \right)+y\left(-c-\lambda \right)+z\left(a\lambda -b\right)=0$$...\left(iv\right)$

Plane $P3$ also passes through the line of intersection of planes $P1$ and $P2$ . Therefore equations $\left(iii\right)$ and $\left(iv\right)$ must be identical

On comparing coefficients we get,

$\frac{1+c\lambda }{b}=\frac{-\left(c+\lambda \right)}{a}=\frac{a\lambda -b}{-1}$

$\therefore$$a\left(1+c\lambda \right)=-b\left(c+\lambda \right)$

$\Rightarrow$ $\lambda =\frac{-\left(a+bc\right)}{ac+b}$ $...\left(v\right)$

$\therefore$ $\left(c+\lambda \right)=a\left(a\lambda -b\right)$

$\Rightarrow$ $\lambda =\frac{c+ab}{a^2-1}$ $...\left(vi\right)$

From $\left(v\right),\left(vi\right)$

$\frac{-\left(a+bc\right)}{ac+b}=\frac{c+ab}{a^2-1}$

$\Rightarrow$ $-a^3-a^{2}bc+a+bc=a{c}^2+a^{2}bc+bc+a{b}^2$

$\Rightarrow$$2a^{2}bc+a{b}^2+a{c}^2+a^3-a=0$

$\Rightarrow$ $2abc+b^2+c^2+a^2-1=0$

$\Rightarrow$ $a^2+b^2+c^2+2abc=1$

So the value of $a^2+b^2+c^2+2abc$ is $1$ for the given equations of planes.

Hence, option (B) is the correct answer.