$\hat{\mathbf{r}} .(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=5$ and $\hat{\mathbf{r}} .(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=3$ are two planes. A plane $\pi$ passing through the line of intersection of these two planes, passes through the point $(0,1,2)$. If the equation of $\pi$ is $\hat{\mathbf{r}} .(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}})=m$, then $\frac{b c}{a^{2}}=$

If $A(-2,4, a), B(1, b, 3), C(c, 0,4)$ and $D(-5,6,1)$ are collinear points, then $a+b+c=$

$A(1,-2,1)$ and $B(2,-1,2)$ are the end points of a line segment. If $D(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from $C(1,2,3)$ to $A B$, then $\alpha^{2}+\beta^{2}+\gamma^{2}=$

The foot of the perpendicular drawn from the point $(-2,-1,3)$ to a plane $\pi$ is $(1,0,-2)$. If $a, b, c$ are the intercepts made by the plane $\pi$ on $X, Y, Z$-axis respectively, then $3 a+b+5 c=$