If $(\alpha, \beta, \gamma)$ is a triad of real numbers satisfying $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}=\alpha(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})+\beta(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})+\gamma(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})$, then $\alpha^2-\beta^2+\gamma^2=$

Let $L$ be a line passing through a point $A$ and parallel to the vector $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$. Let $-7 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}$ be the position vector of a point $P$ on $L$ such that $|\mathbf{A P}|=12$. Then, the position vector of $\mathbf{A}$ can be

If $\theta$ is the angle between the vectors $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $a \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+b \hat{\mathbf{k}}$ and $\cos \theta=\frac{2}{3}$, then $2(a+b+3)=$

A bisector of the angle between the normals of the planes $4 x+3 y=5$ and $x+2 y+2 z=4$ is along the vector