If $\alpha, \beta$ and $\gamma$ are real numebrs such that

$$ \begin{aligned} & \left(\frac{7}{3}+\beta\right) \hat{\mathbf{i}}-\hat{\mathbf{j}}+(\alpha+\gamma) \hat{\mathbf{k}} \\ & =\frac{5}{3}(\alpha \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+\beta(2 \hat{\mathbf{j}}+\hat{\mathbf{k}})+(\hat{\mathbf{i}}+\gamma \hat{\mathbf{j}}+3 \hat{\mathbf{k}}), \text { then } \\ & 5 \alpha-9 \beta+13 \gamma= \end{aligned} $$

If $\mathbf{r}=(2-\lambda+\mu) \hat{\mathbf{i}}+(1-\mu) \hat{\mathbf{j}}+(2-3 \lambda+2 \mu) \hat{\mathbf{k}}$ is the vector equation of a plane, then the equivalent cartesian equation of the plane is

If $\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{x}=\left(\frac{\mathbf{a b}}{|\mathbf{b}|^2}\right) \mathbf{b}, \mathbf{y}=\left(\frac{\mathbf{a b}}{|\mathbf{a}|^2}\right) \mathbf{a}$ and $\theta$ is angle between $\mathbf{a}$ and $\mathbf{b}$, then $x^2+y^2=$

Let $\pi_1$ be a plane passing through the point $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and perpendicular to the vector $-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$. Let the line $L$ passing through the points $3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ be a normal to the plane $\pi_2$. If the angle between the planes $\pi_1$ and $\pi_2$ is $\theta$, then $\cos \theta=$