A plane $\pi$ passing through the points $2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}, 3 \hat{\mathbf{i}}+4 \hat{\mathbf{k}}$ is parallel to the vector $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$. If a line joining the points $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$ and $\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ intersects the plane $\pi$ at the point $a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$, then $a+b+2 c=$

$\hat{\mathbf{r}} .(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=5$ 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 $\alpha^{2}+\beta^{2}+\gamma^{2}=$