Three force $2\mathop i\limits^ \to - 3\mathop j\limits^ \to + 4\mathop k\limits^ \to ,{\text{ 8}}\mathop i\limits^ \to - 7\mathop j\limits^ \to + 6\mathop k\limits^ \to {\text{ }}and{\text{ }}m(\mathop i\limits^ \to - \mathop j\limits^ \to + \mathop k\limits^ \to )$ equilibrate a particle, then m is?
(A) 10
(B) -10
(C) 20
(D) -15

Hint: It is given that the three forces are in equilibrium, when forces are in equilibrium then the sum of that three forces is zero.so, to find the value of m multiply the three forces and equate the resultant value with zero.

Complete step by step answer
Formula used
In equilibrium net force is equal to zero
${\text{Net force = 0}}$
Let us see about what an equilibrium is and the terms related to it.
Equilibrium is defined as the balanced state, an object is said to be in equilibrium when, the forces acting on an object cancel each other to help the object to continue the state of rest or uniform motion.
We can also say, if the size and direction of the forces acting on an object are exactly balanced, then there is no net force acting on the object and the object is said to be in equilibrium.
Principles of Equilibrium:
1. The sum of forces exerted is equal to zero.
2. The sum of torques is equal to zero.
In equilibrium the forces acting on the object cancel each other so the net force acting on the object will be zero.
Given,
The three force equilibrate the particle are $2\mathop i\limits^ \to - 3\mathop j\limits^ \to + 4\mathop k\limits^ \to ,{\text{ 8}}\mathop i\limits^ \to - 7\mathop j\limits^ \to + 6\mathop k\limits^ \to {\text{ }}and{\text{ }}m(\mathop i\limits^ \to - \mathop j\limits^ \to + \mathop k\limits^ \to )$
The net force on the particle is the sum of these three forces
In equilibrium the net force acting on an object is zero
So, Net force = Zero
 \[ \Rightarrow {\text{ }}(2\mathop i\limits^ \to - 3\mathop i\limits^ \to + 4\mathop k\limits^ \to {\text{) + (8}}\mathop i\limits^ \to - 7\mathop j\limits^ \to + 6\mathop k\limits^ \to {\text{) + }}m(\mathop i\limits^ \to - \mathop j\limits^ \to + \mathop k\limits^ \to ){\text{ = 0}}\]
\[ \Rightarrow {\text{ }}(2\mathop i\limits^ \to - 3\mathop i\limits^ \to + 4\mathop k\limits^ \to {\text{) + (8}}\mathop i\limits^ \to - 7\mathop j\limits^ \to + 6\mathop k\limits^ \to {\text{) + }}(m\mathop i\limits^ \to - m\mathop j\limits^ \to + m\mathop k\limits^ \to ){\text{ = 0}}\]
\[ \Rightarrow {\text{ (2}} + 8 + {\text{m)}}\mathop i\limits^ \to {\text{ + (}} - {\text{3}} - {\text{7}} - {\text{m)}}\mathop j\limits^ \to {\text{ + (}}4 + 6 + {\text{m)}}\mathop k\limits^ \to {\text{ = 0}}\]
\[ \Rightarrow {\text{ (10}} + {\text{m )}}\mathop i\limits^ \to {\text{ + (}} - 10 + {\text{m)}}\mathop j\limits^ \to {\text{ + (10}} + {\text{m)}}\mathop k\limits^ \to {\text{ = 0}}\mathop i\limits^ \to + 0\mathop j\limits^ \to + 0\mathop k\limits^ \to \]
Equating both sides we get,
$ \Rightarrow {\text{ }}10 + m = 0{\text{ ; }} - 10 - m = 0{\text{ ; 10}} + m = 0$
$ \Rightarrow {\text{ m = 10}}$

Hence the correct answer is option (A) 10.

Note: The concept of equilibrium was introduced by Galileo in physics. The term equilibrium also exists in chemistry. Equilibrium in chemistry was introduced by Berthollet which deals with the chemical reactions which are reversible at equal rates.