A vector which makes equal angle with the vectors \[\dfrac{1}{3}\left( {i - 2j + 2k} \right)\] , \[\dfrac{1}{4}\left( { - 4i - 3k} \right)\] and $j$ is
A \[5i + j + 5k\]
B $-6i + j + 5k$
C $5i - j - 5k$
D $5i + j - 5k$

Hint: First let us take unknown vector is equal to $\overrightarrow a $ = ${a_1}i + {a_2}j + {a_3}k$ hence it is given that it make equal angle with all vectors hence $\cos \theta = \dfrac{{\overrightarrow a .\vec b}}{{\left| a \right|\left| b \right|}} = \dfrac{{\vec a.\vec c}}{{\left| a \right|\left| c \right|}} = \dfrac{{\vec a.\vec d}}{{\left| a \right|\left| d \right|}}$ now put the value and form three equations and get the value of ${a_1},{a_2},{a_3}$.

Complete step-by-step answer:
As in the question first let us suppose that the unknown vector is equal to $\overrightarrow a $ = ${a_1}i + {a_2}j + {a_3}k$
Hence it is given that this vector ${a_1}i + {a_2}j + {a_3}k$ makes equal angle with the \[\dfrac{1}{3}\left( {i - 2j + 2k} \right)\] , \[\dfrac{1}{4}\left( { - 4i - 3k} \right)\] and $j$ .
$\overrightarrow b $ = \[\dfrac{1}{3}\left( {i - 2j + 2k} \right)\]
 $\overrightarrow c $ = \[\dfrac{1}{4}\left( { - 4i - 3k} \right)\]
 $\overrightarrow d $ = $j$
when we take Dot product as $\overrightarrow a .\overrightarrow b = \left| a \right|\left| b \right|\cos \theta $,
So the $\cos \theta $ is same for everyone because it is given that the vector make equal angle with all other vectors
As $\cos \theta $ is same for everyone then
$\cos \theta = \dfrac{{\overrightarrow a .\vec b}}{{\left| a \right|\left| b \right|}} = \dfrac{{\vec a.\vec c}}{{\left| a \right|\left| c \right|}} = \dfrac{{\vec a.\vec d}}{{\left| a \right|\left| d \right|}}$

As we know that $\left| a \right| = \sqrt {{a_1}^2 + {a_2}^2 + {a_3}^2} $
$\left| b \right| = \dfrac{1}{3}\sqrt {{1^2} + {{( - 2)}^2} + {{(2)}^2}} = 1$
$\left| c \right| = \dfrac{1}{5}\sqrt {{{( - 4)}^2} + {{( - 3)}^2}} = 1$
\[\left| d \right| = \sqrt {{1^2}} = 1\]
 As $\left| b \right| = \left| c \right| = \left| d \right| = 1$
then from the above , $\cos \theta = \dfrac{{\overrightarrow a .\vec b}}{{\left| a \right|}} = \dfrac{{\vec a.\vec c}}{{\left| a \right|}} = \dfrac{{\vec a.\vec d}}{{\left| a \right|}}$
we know that $\overrightarrow a $ = ${a_1}i + {a_2}j + {a_3}k$
$\cos \theta = \dfrac{{({a_1}i + {a_2}j + {a_3}k).\vec b}}{{\left| a \right|}} = \dfrac{{({a_1}i + {a_2}j + {a_3}k).\vec c}}{{\left| a \right|}} = \dfrac{{({a_1}i + {a_2}j + {a_3}k).\vec d}}{{\left| a \right|}}$
Now by putting the value of $\overrightarrow b $ = \[\dfrac{1}{3}\left( {i - 2j + 2k} \right)\] , $\overrightarrow c $ = \[\dfrac{1}{4}\left( { - 4i - 3k} \right)\] , $\overrightarrow d $ = $j$
$\cos \theta = \dfrac{{({a_1}i + {a_2}j + {a_3}k).\dfrac{1}{3}(i - 2j + 2k)}}{{\left| a \right|}} = \dfrac{{({a_1}i + {a_2}j + {a_3}k).\dfrac{1}{5}( - 4i - 3k)}}{{\left| a \right|}} = \dfrac{{({a_1}i + {a_2}j + {a_3}k).j}}{{\left| a \right|}}$
Multiple $\left| a \right|$ in whole equation
\[\left| a \right|\cos \theta = \dfrac{{{a_1} - 2{a_2} + 2{a_3}}}{3} = \dfrac{{ - 4{a_1} - 3{a_3}}}{5} = \dfrac{{{a_2}}}{1}\]
Now just take \[\dfrac{{{a_1} - 2{a_2} + 2{a_3}}}{3} = \dfrac{{{a_2}}}{1}\] from cross multiplication we get ,
\[{a_1} - 2{a_2} + 2{a_3} = 3{a_2}\]
\[{a_1} - 5{a_2} + 2{a_3} = 0\]
Now take \[\dfrac{{ - 4{a_1} - 3{a_3}}}{5} = \dfrac{{{a_2}}}{1}\]
  \[ - 4{a_1} - 3{a_3} = 5{a_2}\]
Put the value of $5{a_2}$ in the equation \[{a_1} - 5{a_2} + 2{a_3} = 0\]
\[{a_1} + 4{a_1} + 3{a_3} + 2{a_3} = 0\] or \[5{a_1} + 5{a_3} = 0\]
Now solving further we get \[{a_1} = - {a_3}\]
Now put the value of \[{a_1} = - {a_3}\] in the equation \[{a_1} - 5{a_2} + 2{a_3} = 0\]
\[ - {a_3} - 5{a_2} + 2{a_3} = 0\] or \[{a_2} = \dfrac{{{a_3}}}{5}\]
Hence we got the value of ${a_1}$ and ${a_2}$ in the term of ${a_3}$
Initially our equation of vector is ${a_1}i + {a_2}j + {a_3}k$ by putting the value of ${a_1}$ , ${a_2}$ in the term of ${a_3}$.
we get ,
$ - {a_3}i + \dfrac{{{a_3}}}{5}j + {a_3}k$
As we know that by the multiplication in vector their is no change in it only the magnitude will change that doesn't matters hence
On dividing with ${a_3}$ and multiple by $ - 5$ we get
$5i - j - 5k$

So, the correct answer is “Option C”.

Note: Always being careful at \[\left| a \right|\cos \theta = \dfrac{{{a_1} - 2{a_2} + 2{a_3}}}{3} = \dfrac{{ - 4{a_1} - 3{a_3}}}{5} = \dfrac{{{a_2}}}{1}\] this step for finding the value ${a_1}$ , ${a_2}$ in the term of ${a_3}$.
We will also find the solution of this by equating it equal to any constant term as k . \[k = \dfrac{{{a_1} - 2{a_2} + 2{a_3}}}{3} = \dfrac{{ - 4{a_1} - 3{a_3}}}{5} = \dfrac{{{a_2}}}{1}\] Now make three equation in three variable as ${a_1}$ , ${a_2}$ , ${a_3}$ find the value of ${a_1}$ , ${a_2}$ , ${a_3}$ in the term of k.