The position vectors of two points $A$ and $B$ are $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $7 \hat{\mathbf{i}}-\hat{\mathbf{k}}$ respectively. The point $P$ with position vector $-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ is on the line $A B$. If the point $Q$ is the harmonic conjugate of $P$, then the sum of the scalar components of the position vector of $Q$ is

If $\mathbf{a}$ and $\mathbf{b}$ are two vectors such that $|\mathbf{a}|=5,|\mathbf{b}|=12$ and $|\mathbf{a}-\mathbf{b}|=13$, then $|2 \mathbf{a}+\mathbf{b}|=$

If $\mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are two vectors, then $(\mathbf{a}+2 \mathbf{b}) \times(3 \mathbf{a}-\mathbf{b})$