Answer
$\text{Scalar;}-72$.
Work Step by Step
Put $v=6\mathbf{i}-\mathbf{j}$ and $w=-3\mathbf{i}$ in the expression ${{\left\| v+w \right\|}^{2}}-{{\left\| v-w \right\|}^{2}}$. Then, we have
$\begin{align}
& {{\left\| v+w \right\|}^{2}}-{{\left\| v-w \right\|}^{2}}={{\left\| 6\mathbf{i}-\mathbf{j}+\left( -3\mathbf{i} \right) \right\|}^{2}}-\left\| 6\mathbf{i}-\mathbf{j}-\left( -3\mathbf{i} \right) \right\| \\
& ={{\left\| 6\mathbf{i}-\mathbf{j}-3\mathbf{i} \right\|}^{2}}-{{\left\| 6\mathbf{i}-\mathbf{j}+3\mathbf{i} \right\|}^{2}} \\
& ={{\left\| 3\mathbf{i}-\mathbf{j} \right\|}^{2}}-{{\left\| 9\mathbf{i}-\mathbf{j} \right\|}^{2}}
\end{align}$
If $u=a\mathbf{i}+b\mathbf{j}$ is a vector, then its norm is $\left\| u \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$. Use this definition of norm to get
$\begin{align}
& {{\left\| v+w \right\|}^{2}}-{{\left\| v-w \right\|}^{2}}={{3}^{2}}+{{\left( -1 \right)}^{2}}-\left( {{9}^{2}}+{{\left( -1 \right)}^{2}} \right) \\
& =9+1-\left( 81+1 \right) \\
& =10-82 \\
& =-72
\end{align}$
Hence, the given expression is a scalar and its value is $-72$.
