If $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \mathbf{c}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$. $\mathbf{d}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ are four vector, then $(\mathbf{a} \times \mathbf{c}) \times(\mathbf{b} \times \mathbf{d})=$

The angle between the diagonals of the parallelogram whose adjacent sides are $2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}, \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ is

If the points having the position vectors $-i+4 j-4 k_{\text {, }}$, $3 i+2 j-5 k,-3 i+8 j-5 k$ and $-3 i+2 j+\lambda k$ are coplanar, then $\lambda=$

If $|f|=10,|g|=14$ and $|f-g|=15$, then $|f+g|=$