If $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-7}{2}$ lies in the plane $a x+b y+z=7$, then $a+b=$

If the point of intersection of the lines $\mathbf{r}=\hat{\mathbf{i}}-6 \hat{\mathbf{j}}+(p \sec \alpha) \hat{\mathbf{k}}+t(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})$ and $\mathbf{r}=4 \hat{\mathbf{j}}+\hat{\mathbf{k}}+\lambda(2 \hat{\mathbf{i}}+(p \tan \alpha) \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$ is $8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+9 \hat{\mathbf{k}}$, (where $\left.0<\alpha<\frac{\pi}{2}\right)$, then $p=$

$\mathbf{1}, \mathbf{m}, \mathbf{n}$ are three unit vectors in a right handed system and $L$ is a line through the points $A, B, C$ whose position vectors are $p \mathbf{1}+7 \mathbf{m}-6 \mathbf{n}, 2 \mathbf{1}+5 \mathbf{m}-4 \mathbf{n}$ and $1+4 \mathbf{m}-3 \mathbf{n}$ respectively. If the equation of the plane containing $L$ and the points ( $-p, p, p+1$ ) is $a x+b y+c z=1$, then $p(a+b+c)=$

$E(1,0,0), F(0,2,0), G(0,0,3)$ are respectively the mid-points of the sides $A B, B C, C A$ of $\triangle A B C$. If $a_1, b_1, c_1$ and $a_2, b_2, c_2$ are respectively the direction ratios of $A F$ and $B G$, then $\frac{a_1^2+b_1^2+c_1^2}{a_2^2+b_2^2+c_2^2}=$