Unit vectors equally inclined to the vectors $\hat i,\frac{1}{3}(-2\hat i+\hat j+2\hat k), -\frac{1}{5}(4\hat j+3\hat k)$ are

$±\frac{1}{\sqrt{51}}(\hat i-5\hat j+5\hat k)$

Let the require unit vector be $\vec b = x\hat i+y\hat j+z\hat k$.

It is equally inclined to the given units vectors. Therefore,

$(x\hat i+y\hat j+z\hat k). i =\frac{1}{3} (-2\hat i+\hat j+2\hat k). (x\hat i + y\hat j +z\hat k)$

$=-\frac{1}{5}(4\hat j+3\hat k). (x\hat i + y\hat j +3\hat k)$

$⇒x=\frac{1}{3}(-2x+y+2z)=-\frac{1}{5}(4y+ 3z)$

$⇒5x-y-2z=0$ and $5x + 4y + 3z = 0$

$⇒\frac{x}{5}=\frac{y}{-25}=\frac{z}{25}$

$⇒\frac{x}{1}=\frac{y}{-5}=\frac{z}{5}=λ(say) ⇒ x=λ, y=-5λ, z=5λ$

It is given that

$\vec r=x\hat i+y\hat j+z\hat k$ is a unit vector.

$∴|\vec r|=1⇒x^2 + y^2+z^2=1⇒ λ=±\frac{1}{\sqrt{51}}$

$∴\vec r=±\frac{1}{\sqrt{51}}(\hat i-5\hat j+5\hat k)$