Let A(alpha, 4, 7) and B(3, beta, 8) be two p | vedclass

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(D) The $YZ$ plane $(x=0)$ divides $AB$ in ratio $2:3$. Using the section formula,the $x$-coordinate of the division point is $\frac{2(3) + 3(\alpha)}

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$\left(\frac{-16}{3}, 10, \frac{19}{3}\right)$

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(D) The $YZ$ plane $(x=0)$ divides $AB$ in ratio $2:3$. Using the section formula,the $x$-coordinate of the division point is $\frac{2(3) + 3(\alpha)}{2+3} = 0$,which gives $6 + 3\alpha = 0$,so $\alpha = -2$. The $ZX$ plane $(y=0)$ divides $AB$ in ratio $4:5$. Using the section formula,the $y$-coordinate of the division point is $\frac{4(\beta) + 5(4)}{4+5} = 0$,which gives $4\beta + 20 = 0$,so $\beta = -5$. We need to find the point $C$ that divides $\overline{AB}$ externally in the ratio $\alpha : \beta = -2 : -5$,which is equivalent to $2:5$ externally. The external division formula for points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ in ratio $m:n$ is $\left(\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}, \frac{mz_2 - nz_1}{m-n}\right)$. Substituting $A(-2, 4, 7)$,$B(3, -5, 8)$,$m=2$,and $n=5$: $x = \frac{2(3) - 5(-2)}{2-5} = \frac{6+10}{-3} = -\frac{16}{3}$. $y = \frac{2(-5) - 5(4)}{2-5} = \frac{-10-20}{-3} = \frac{-30}{-3} = 10$. $z = \frac{2(8) - 5(7)}{2-5} = \frac{16-35}{-3} = \frac{-19}{-3} = \frac{19}{3}$. Thus,the point $C$ is $\left(-\frac{16}{3}, 10, \frac{19}{3}\right)$.

Let $A(\alpha, 4, 7)$ and $B(3, \beta, 8)$ be two points in space. If the $YZ$ plane and $ZX$ plane respectively divide the line segment joining the points $A$ and $B$ in the ratio $2:3$ and $4:5$,then the point $C$ which divides $\overline{AB}$ in the ratio $\alpha: \beta$ externally is

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