If the foot of the perpendicular drawn from the point $(2,0,-3)$ to the plane $\pi$ is $(1,-2,0)$ and the equation of the plane $\pi$ is $a x+b y-3 z+d=0$, then $a+b+d=$

Let $\pi_1$ be the plane determined by the vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}$. $\hat{\mathbf{i}}+\hat{\mathbf{k}}$ and $\pi_2$ be the plane determined by the vectors $\hat{\mathbf{j}}-\hat{\mathbf{k}}, \hat{\mathbf{k}}-\hat{\mathbf{i}}$. Let $\mathbf{a}$ be a non-zero vector parallel to the line of intersection of the planes $\pi_1$ and $\pi_2$. If $\mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$, then the angle between the vectors $\mathbf{a}$ and $\mathbf{b}$ is

If $m: n$ is the ratio in which the point $\left(\frac{8}{5},-\frac{1}{5}, \frac{8}{5}\right)$ divides the segment joining the points $(2, p, 2)$ and $(p,-2, p)$, where $p$ is an integer than $\frac{3 m+n}{3 n}=$

If $(\alpha, \beta \gamma)$ is the foot of the perpendicular drawn from a point $(-1,2,-1)$ to the line joining the points $(2,-1,1)$ and ( $1,1-2$ ), then $\alpha+\beta+\gamma=$