Step 1: Understand the Section Formula for External Division.
The section formula for external division is given by: \[ \frac{x_2 - x_1}{y_2 - y_1} = \frac{m}{n} \] where \(A(x_1, y_1, z_1)\), \(B(x_2, y_2, z_2)\), and \(C(x, y, z)\) are the points involved, and \(C\) divides the line segment \(AB\) externally in the ratio \(m:n\).
Step 2: Apply the formula to the given points.
We are given: \[ A = (1, 2, 3), \quad B = (3, 4, 7), \quad C = (-3, -2, -5). \] We will apply the section formula for the external division of the line segment to find the ratio in which \(C\) divides \(AB\) externally.
Step 3: Use the formula to find the ratio.
The section formula for external division gives: \[ \frac{x_2 - x_1}{x - x_1} = \frac{m}{n}, \quad \frac{y_2 - y_1}{y - y_1} = \frac{m}{n}, \quad \frac{z_2 - z_1}{z - z_1} = \frac{m}{n}. \] Substituting the values of the coordinates of \(A\), \(B\), and \(C\), we calculate the ratio.
Step 4: Final Answer.
The ratio in which \(C\) divides \(AB\) externally is \(2:3\).
Given $\triangle ABC \sim \triangle PQR$, $\angle A = 30^\circ$ and $\angle Q = 90^\circ$. The value of $(\angle R + \angle B)$ is
Which of the following are ambident nucleophiles?
[A.] CN$^{\,-}$
[B.] CH$_{3}$COO$^{\,-}$
[C.] NO$_{2}^{\,-}$
[D.] CH$_{3}$O$^{\,-}$
[E.] NH$_{3}$
Identify the anomers from the following.

The standard Gibbs free energy change \( \Delta G^\circ \) of a cell reaction is \(-301 { kJ/mol}\). What is \( E^\circ \) in volts?
(Given: \( F = 96500 { C/mol}\), \( n = 2 \))