If the resolved part of the force vector \[5\widehat i + 4\widehat j + 2\widehat k\] along and perpendicular to the vector\[3\widehat i + 4\widehat j + - 5\widehat k\] are \[\alpha \] and \[\beta \] respectively. Then the value of \[\alpha \] is
a.\[\dfrac{{21}}{{50}}\left( {3\widehat i + 4\widehat j - 5\widehat k} \right)\]
b.\[\dfrac{{21}}{{50}}\left( {3\widehat i - 4\widehat j + 5\widehat k} \right)\]
c.\[\dfrac{{11}}{{50}}\left( {2\widehat i - 4\widehat j + 3\widehat k} \right)\]
d.\[\dfrac{1}{{50}}\left( {187\widehat i + 116\widehat j + 205\widehat k} \right)\]

Hint: Here we need to find the value of \[\alpha \], which is equal to the resolved part of vector \[5\widehat i + 4\widehat j + 2\widehat k\] along the vector \[3\widehat i + 4\widehat j + - 5\widehat k\]. We will resolve the vector \[5\widehat i + 4\widehat j + 2\widehat k\] along another vector\[3\widehat i + 4\widehat j + - 5\widehat k\]. We will use the formula to resolve the vector\[5\widehat i + 4\widehat j + 2\widehat k\] along another vector\[3\widehat i + 4\widehat j + - 5\widehat k\]. This will give us the value of \[\alpha \].

Formula used:
 We will use the formula to resolve the part of any vector \[a\] along another vector \[b\] is \[\dfrac{{\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow b }}{{{{\left| b \right|}^2}}}\].

Complete step-by-step answer:
It is given that the resolved part of the vector \[5\widehat i + 4\widehat j + 2\widehat k\] along the vector \[3\widehat i + 4\widehat j + - 5\widehat k\] is equal to \[\alpha \]and resolved part of the vector \[5\widehat i + 4\widehat j + 2\widehat k\]perpendicular to the vector\[3\widehat i + 4\widehat j + - 5\widehat k\] is equal to\[\beta \].
Let us assume \[\overrightarrow a = 5\widehat i + 4\widehat j + 2\widehat k\] and \[\overrightarrow b = 3\widehat i + 4\widehat j + - 5\widehat k\].
We will put the value of the vector\[\overrightarrow a \] and \[\overrightarrow b \] in the formula $\Rightarrow$ \[\dfrac{{\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow b }}{{{{\left| b \right|}^2}}}\].
$\Rightarrow$ \[\alpha = \dfrac{{\left[ {\left( {5\widehat i + 4\widehat j + 2\widehat k} \right).\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)} \right]\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)}}{{{{\left| {\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)} \right|}^2}}}\]
We find the dot product of the two vectors in the numerator.
$\Rightarrow$ \[\alpha = \dfrac{{\left( {5.3 + 4.4 - 2.5} \right)\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)}}{{{{\left| {\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)} \right|}^2}}}\]
Multiplying and adding the terms inside the bracket, we get
$\Rightarrow$ \[\alpha = \dfrac{{21\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)}}{{{{\left| {\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)} \right|}^2}}}\]
Now, we will find the magnitude of the vector in the denominator.
$\Rightarrow$ \[\alpha = \dfrac{{21\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)}}{{{{\left| {\sqrt {\left( {{3^2} + {4^2} + {{\left( { - 5} \right)}^2}} \right)} } \right|}^2}}}\]
Simplifying the expression, we get
$\Rightarrow$ \[\alpha = \dfrac{{21\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)}}{{50}}\]
Rewriting the value of \[\alpha \], we get
$\Rightarrow$ \[\alpha = \dfrac{{21}}{{50}}\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)\]
Thus, the correct option is option (a).

Note: We have calculated the resolved part of a vector along another vector. Vector resolution is defined as a process of breaking one vector into two smaller vectors where both the vectors will be perpendicular to each other. In other words, we can say that the vector resolution is a process of finding two components of a given vector.