Given,

u=(-6,4,2) v=(3,1,5)

To find : - A vector orthogonal to both u=(-6,4,2) and v=(3,1,5).

Let $(u_{1}, u_{2}, u_{3})$ be the required unit vector. Then we have,

$\implies -6u_{1} + 4u_{2} + 2u_{3} = 0$

$\implies 3u_{1} + u_{2} + 5u_{3} = 0$

Solving these equations,

$$\begin{bmatrix} -6 & 4 & 2 \\ 3 & 1 & 5 \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\u_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\ R_{2} \rightarrow 2R_{2} + R_{1} \\ \begin{bmatrix} -6 & 4 & 2 \\ 0 & 6 & 12 \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0\end{bmatrix} \\ R_{1} \rightarrow 6R_{1} - R_{2} \\ \begin{bmatrix} -36 & 18 & 0 \\ 0 & 6 & 12\end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3}\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ $\implies -36u_{1} + 18u_{2} = 0$

$\implies Means, u_{2} = 2u_{1}$

$\implies 6u_{2} + 12u_{3} = 0$

$\implies Means, u_{2} + 2u_{3} = 0$

Let $u_{2} = y$

$\therefore u_{1} = \frac{t}{2}$ & $u_{3} = -\frac{t}{2}$

Put t = 2

$\therefore u_{1} = 1, u_{2} = 2, u_{3} = -1$

Now $\sqrt{u_{1}^2 + u_{2}^2 + u_{3}^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$

$\therefore\ Required\ vector\ is\ \frac{1}{\sqrt{6}} (1, 2, -1). Ans.$