If \[\overrightarrow{a}\overrightarrow{,b},\overrightarrow{c}\] are mutually perpendicular vectors of equal magnitude, then \[\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\] is equally inclined to
(a). \[\overrightarrow{a}\] only
(b). \[\overrightarrow{a},\overrightarrow{c}\] only
(c). All \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\]
(d). None of these

Hint: First of all we have to determine the relation between the vectors \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] as in the mutually perpendicular case. Then we have to assume that \[\left| \overrightarrow{a} \right|=\left| \overrightarrow{b} \right|=\left| \overrightarrow{c} \right|\] which is equal to its constant. Then we have to determine the sum of the square for \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\]. After that you have to use the formula for finding the angle between the vectors \[\overrightarrow{a}\] and \[\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right),\overrightarrow{b}\] and \[\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)\] and \[c\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)\]. Now use can easily find out the vector which is equally inclined.

Complete step by step solution:
In the given question we have given that vectors are equal in magnitudes \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] are of equal magnitudes
So, \[\left| \overrightarrow{a} \right|=\left| \overrightarrow{b} \right|=\left| \overrightarrow{c} \right|\]
Also as per given in the question \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] are mutually perpendicular to each other.
So, \[\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0\]
Now we know that
\[\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right).\overrightarrow{a}=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{a} \right|\cos \alpha \] Where \[\alpha =angle\] between \[\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)\] and \[\overrightarrow{a}\].
Now we will open the bracket of the light hand side and the resulted equation will be,
\[\overrightarrow{a},\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{a}+\overrightarrow{c}.\overrightarrow{a}=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{a} \right|\cos \alpha .............\left( i \right)\]
Now we will use the property of vectors
Property: \[\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{a}\] and \[\overrightarrow{a},\overrightarrow{a}={{\left| \overrightarrow{a} \right|}^{2}}\] apply the properties on equation (i)
\[\Rightarrow {{\left| \overrightarrow{a} \right|}^{2}}+0+0=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{a} \right|\cos \alpha \]
Now we will find the value of \[\cos \alpha \] from the above equation
\[\Rightarrow \cos \alpha =\dfrac{\left| \overrightarrow{a} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}\]
Now we will take
\[\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right),\overrightarrow{b}=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{b} \right|\cos \beta \] Where \[\beta =angle\] between \[\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)\] and \[\overrightarrow{b}\].
Now we will open the bracket of the left-handed side and the resulted equation will be,
\[\Rightarrow \overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{b}+\overrightarrow{c}.\overrightarrow{b}=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{b} \right|\cos \beta ........(ii)\]
Now we will again use the property of vectors
Property: \[\overrightarrow{a},\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{a}\] and \[\overrightarrow{a}.\overrightarrow{a}={{\left| \overrightarrow{a} \right|}^{2}}\] apply these properties on equation (ii)
\[\Rightarrow 0+\left| {{\overrightarrow{b}}^{2}} \right|+0=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{b} \right|\cos \beta \]
Now we will find the value of \[\cos \beta \] from the above equation
\[\Rightarrow \cos \beta =\dfrac{\left| \overrightarrow{b} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}\]
Now we will take
\[\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{b} \right).\overrightarrow{c}=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{b} \right|\cos \gamma \] Where \[\gamma =\] angle between \[\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)\] and \[\overrightarrow{c}\].
Now we will open the bracket of the left hand side and the resulted equation will be,
\[\Rightarrow \overrightarrow{a}.\overrightarrow{c}+\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{c}=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{c} \right|\cos \gamma .........(iii)\]
Now we will again use the property of vectors
Property: \[\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{a}\] and \[\overrightarrow{a}.\overrightarrow{a}={{\left| \overrightarrow{a} \right|}^{2}}\] apply the perpendicular on equation (iii)
\[\Rightarrow 0+0+{{\left| \overrightarrow{c} \right|}^{2}}=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{c} \right|\cos \gamma \]
Now we will find the value of \[\cos \gamma \] from the above equation
\[\Rightarrow \cos \gamma =\dfrac{\left| \overrightarrow{c} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}\]
As calculated above in the question
\[\Rightarrow \cos \alpha =\dfrac{\left| \overrightarrow{a} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|},\cos \beta =\dfrac{\left| \overrightarrow{b} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|},\cos \gamma =\dfrac{\left| \overrightarrow{c} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}\]
According to the information given in the question we know the
\[\Rightarrow \left| \overrightarrow{a} \right|=\left| \overrightarrow{b} \right|=\left| \overrightarrow{c} \right|\]
Thus we can say that \[\cos \alpha =\cos \beta =\cos \gamma \] and according to the above equation we can say that
\[\Rightarrow \alpha =\beta =\gamma \]
Therefore, \[\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)\] is equation inclined to \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\].

So, the correct answer is “Option c”.

Note: We know that the dot product in between \[2\] unit vectors is very simple to compute. If the vector is length type then we have \[3\] vector of dot product i.e. \[\overrightarrow{i}.\overrightarrow{j},\overrightarrow{k}\]. It becomes, \[\overrightarrow{i},\overrightarrow{i}=\overrightarrow{j}.\overrightarrow{j}=\overrightarrow{k}.\overrightarrow{k}=1\] if vector are of length \[\overrightarrow{a}\] then, we get \[\overrightarrow{i},\overrightarrow{i}=\overrightarrow{j}.\overrightarrow{j}=\overrightarrow{k}.\overrightarrow{k}=a\].