The point which lies on the plane passing through the point $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, 3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}},-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ is
If $\mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$, then the component of $\mathbf{b}$ perpendicular to $\mathbf{a}$ is
If the angle between the planes $\mathbf{r} \cdot(11 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\alpha \hat{\mathbf{k}})=7$ and $\mathbf{r} \cdot(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})=5$ is $\frac{\pi}{2}$, then $\alpha=$
If $2 \mathbf{i}+3 \mathbf{j}-4 \mathbf{k}$ and $-\mathbf{i}+2 \mathbf{j}+\mathbf{k}$ are the two diagonals of a parallelogram, then the area of the parallelogram in square units is
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