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# If the planes x – cy – bz = 0, cx – y + az = 0 and bx + ay – z = 0, pass through a line, then find the value of ${a^2} + {b^2} + {c^2} + 2abc$ is$\left( A \right)$ -1$\left( B \right)$ 2$\left( C \right)$ 1$\left( D \right)$ 0  Hint – In this particular type of question use the concept that if the three planes passing through the same line then the intersection of any two places is similar to the third plane and the intersection of two plane is given as, (ax + by + cz + $\lambda$(px + qy + rz)) = 0, where ($\lambda$) is any real parameter, so use these concepts to reach the solution of the question.

Complete step-by-step solution -
Given planes are:
x – cy – bz = 0.................. (1)
cx – y + az = 0.................. (2)
bx + ay – z = 0.................. (3)
Now these three planes passes through the same line so the intersection of equation (1) and equation (2) is same as of the plane 3 (i.e. equation (3))
I.e. the intersection of planes (1) and (2) is similar to the plane (3).
So first find out the intersection of planes (1) and (2) so we have,
The intersection of planes (1) and (2) is given as
$\Rightarrow$ (x – cy – bz) + $\lambda$(cx – y + az) = 0
Now separate x, y and z terms we have,
$\Rightarrow x\left( {1 + c\lambda } \right) - y\left( {c + \lambda } \right) + z\left( { - b + a\lambda } \right) = 0$ ................ (4)
Now equation (3) and (4) are similar planes so the solution of these planes is given according to property,
As we know that the solution of the parallel lines (i.e. slope is same) or the same line (for example ax + by = c and lx + my = n) is given as $\dfrac{a}{l} = \dfrac{b}{m} = \dfrac{c}{n}$ so use this property in equation (3) and (4) the solution is given as,
$\Rightarrow \dfrac{{1 + c\lambda }}{b} = \dfrac{{ - \left( {c + \lambda } \right)}}{a} = \dfrac{{ - b + a\lambda }}{{ - 1}}$.................... (5)
Now we have to eliminate ($\lambda$).
$\Rightarrow \dfrac{{1 + c\lambda }}{b} = \dfrac{{ - \left( {c + \lambda } \right)}}{a}$
$\Rightarrow a + ac\lambda = - bc - b\lambda$
$\Rightarrow b\lambda + ac\lambda = - bc - a$
$\Rightarrow \lambda \left( {b + ac} \right) = - bc - a$
$\Rightarrow \lambda = \dfrac{{ - bc - a}}{{b + ac}}$
Now from equation (5) take last two terms we have,
$\Rightarrow \dfrac{{ - \left( {c + \lambda } \right)}}{a} = \dfrac{{ - b + a\lambda }}{{ - 1}}$
Now substitute the value of ($\lambda$) in the above equation we have,
$\Rightarrow \dfrac{{ - \left( {c + \left( {\dfrac{{ - bc - a}}{{b + ac}}} \right)} \right)}}{a} = \dfrac{{ - b + a\left( {\dfrac{{ - bc - a}}{{b + ac}}} \right)}}{{ - 1}}$
Now simplify this we have,
$\Rightarrow c + \left( {\dfrac{{ - bc - a}}{{b + ac}}} \right) = - ba + {a^2}\left( {\dfrac{{ - bc - a}}{{b + ac}}} \right)$
Now take (b + ac) as LCM we have,
$\Rightarrow \dfrac{{c\left( {b + ac} \right) - bc - a}}{{b + ac}} = \dfrac{{ - ba\left( {b + ac} \right) + {a^2}\left( { - bc - a} \right)}}{{b + ac}}$
$\Rightarrow c\left( {b + ac} \right) - bc - a = - ba\left( {b + ac} \right) + {a^2}\left( { - bc - a} \right)$
$\Rightarrow bc + a{c^2} - bc - a = - a{b^2} - {a^2}bc - {a^2}bc - {a^3}$
Now cancel out the common terms we have,
$\Rightarrow a{c^2} - a = - a{b^2} - {a^3} - 2{a^2}bc$
Now take a common from both sides and cancel out we have,
$\Rightarrow {c^2} - 1 = - {b^2} - {a^2} - 2abc$
$\Rightarrow {a^2} + {b^2} + {c^2} + 2abc = 1$
So this is the required answer.
Hence option (C) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that the solution of the parallel lines (i.e. slope is same) or the same line (for example ax + by = c and lx + my = n) is given as $\dfrac{a}{l} = \dfrac{b}{m} = \dfrac{c}{n}$, so first find out the intersection of two planes as above then using this property solve these equations as above we will get the required answer.

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