The equation of the plane passing through the point of intersection of the planes $2x-y+z-3=0$ and $4x-3y+5z+9=0$ and parallel to the line $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z-3}{5}$ is $\alpha x+\beta y+\gamma z+d=0$. Then $\alpha+\beta+\gamma+d=$

The length of the projection of the line segment joining the points $(5,-1,4)$ and $(4,-1,3)$ on the plane $x+y+z=7$ is

The plane through the intersection of planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to $Y$-axis also passes through the point

Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=7$ and $\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})=9$ and passing through the point $(2,1,3).$

The equation of the plane passing through the point $\hat{i}+2 \hat{j}-\hat{k}$ and perpendicular to the line of intersection of the planes $r \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $r \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ is:

$A$ line $l$ passing through the origin is perpendicular to the lines $l_{1}: \overrightarrow{r}=(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k}$ and $l_{2}: \overrightarrow{r}=(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k}$. If the coordinates of the point in the first octant on $l_{2}$ at a distance of $\sqrt{17}$ from the point of intersection of $l$ and $l_{1}$ are $(a, b, c)$,then $18(a+b+c)$ is equal to ........ .