Explanation:
C The plane passes through the points \((-1,1,1)\)
and \((1,-1,1)\)
\(\therefore\) Equation of plane passing through \((-1,1,1)\) is -
\(\mathrm{a}(\mathrm{x}+1)+\mathrm{b}(\mathrm{y}-1)+\mathrm{c}(\mathrm{z}-1)=0\)
And, when it is passing through \((1,-1,1)\)
\(\therefore a(1+1)+b(-1-1)+c(1-1)=0\)
\(2 a-2 b+0 c=0\)
If this plane is perpendicular to the plane \(x+2 y+2 z=\) 5 then,
\(a+2 b+2 c=0\)
\(\mathrm{Eq}^{\mathrm{n}}\). (ii) and (iii) are identical.
\(\therefore \quad \frac{\mathrm{a}}{-4-0}=\frac{-\mathrm{b}}{4-0}=\frac{\mathrm{c}}{4+2}\)
\(\frac{\mathrm{a}}{-4}=\frac{\mathrm{b}}{-4}=\frac{\mathrm{c}}{6}\)
\(\frac{\mathrm{a}}{-2}=\frac{\mathrm{b}}{-2}=\frac{\mathrm{c}}{3}=\lambda \quad \text { (say) }\)
\(\mathrm{a}=-2 \lambda, \mathrm{b}=-2 \lambda, \mathrm{c}=3 \lambda\)
On putting the values of \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) in eq. (i), we get -
\(-2 \lambda(x+1)-2 \lambda(y-1)+3 \lambda(z-1)=0\)
\(\lambda[-2 x-2-2 y+2+3 z-3]=0\)
\(-2 x-2 y+3 z-3=0\)
\(2 x+2 y-3 z+3=0\)