Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x $-$ 3y + 5z = 8. If the mirror image of the point $\left( {2, - {1 \over 2},2} \right)$ in the rotated plane is B(a, b, c), then :

<p>Consider the equation of plane,</p> <p>$P:(2x + 3y + z + 20) + \lambda (x - 3y + 5z - 8) = 0$</p> <p>$P:(2 + \lambda )x + 3(3 - 3\lambda )y + 1(1 + 5\lambda )z + (20 - 8\lambda ) = 0$</p> <p>$\because$ Plane P is perpendicular to $2x + 3y + z + 20 = 0$</p> <p>So, $4 + 2\lambda + 9 - 9\lambda + 1 + 5\lambda = 0$</p> <p>$\Rightarrow \lambda = 7$</p> <p>$P:9x - 18y + 36z - 36 = 0$</p> <p>or $P:x - 2y + 4z = 4$</p> <p>If image of $\left( {2, - {1 \over 2},2} \right)$ in plane P is (a, b, c) then</p> <p>${{a - 2} \over 1} = {{b + {1 \over 2}} \over { - 2}} = {{c - 2} \over 4}$</p> <p>and $$\left( {{{a + 2} \over 2}} \right) - 2\left( {{{b - {1 \over 2}} \over 2}} \right) + 4\left( {{{c + 2} \over 2}} \right) = 4$$</p> <p>Clearly $a = {4 \over 3}$, $b = {5 \over 6}$ and $c = - {2 \over 3}$</p> <p>So, $a:b:c = 8:5: - 4$</p>