The ratio in which the sphere {x^2} + {y^2} + | vedclass

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(A) Let the ratio in which the sphere divides the line segment $AB$ be $\lambda : 1$. Using the section formula,the coordinates of the point $P$ on th

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$2:3$ externally

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(A) Let the ratio in which the sphere divides the line segment $AB$ be $\lambda : 1$. Using the section formula,the coordinates of the point $P$ on the line segment $AB$ are given by: $P = \left( \frac{27\lambda + 12}{\lambda + 1}, \frac{-9\lambda - 4}{\lambda + 1}, \frac{18\lambda + 8}{\lambda + 1} ight)$. Since the point $P$ lies on the sphere ${x^2} + {y^2} + {z^2} = 504$,we substitute these coordinates into the sphere equation: $\left( \frac{27\lambda + 12}{\lambda + 1} ight)^2 + \left( \frac{-9\lambda - 4}{\lambda + 1} ight)^2 + \left( \frac{18\lambda + 8}{\lambda + 1} ight)^2 = 504$. Factoring out the common terms: $\frac{1}{(\lambda + 1)^2} [ (3(9\lambda + 4))^2 + (-(9\lambda + 4))^2 + (2(9\lambda + 4))^2 ] = 504$. $\frac{(9\lambda + 4)^2}{(\lambda + 1)^2} [ 9 + 1 + 4 ] = 504$. $\frac{(9\lambda + 4)^2}{(\lambda + 1)^2} 	imes 14 = 504$. $\frac{(9\lambda + 4)^2}{(\lambda + 1)^2} = 36$. Taking the square root on both sides: $\frac{9\lambda + 4}{\lambda + 1} = \pm 6$. Case $1$: $9\lambda + 4 = 6\lambda + 6 \implies 3\lambda = 2 \implies \lambda = 2/3$ (Internal division). Case $2$: $9\lambda + 4 = -6\lambda - 6 \implies 15\lambda = -10 \implies \lambda = -2/3$ (External division). Since the option $2:3$ externally is provided,the correct answer is $A$.

The ratio in which the sphere ${x^2} + {y^2} + {z^2} = 504$ divides the line segment $AB$ joining the points $A(12, -4, 8)$ and $B(27, -9, 18)$ is given by

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