The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ is

The distance between the line $\frac{x - 1}{3} = \frac{y + 2}{-2} = \frac{z - 1}{2}$ and the plane $2x + 2y - z = 6$ is

The vector equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$,which is perpendicular to the plane $x-y+z=0$,is

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 ratio in which the plane $\bar{r} \cdot (\hat{i}-2 \hat{j}+3 \hat{k})=17$ divides the line joining the points $-2 \hat{i}+4 \hat{j}+7 \hat{k}$ and $3 \hat{i}-5 \hat{j}+8 \hat{k}$ is:

If the equation of the line passing through the point $(0, -\frac{1}{2}, 0)$ and perpendicular to the lines $\overrightarrow{r} = \lambda(\hat{i} + a\hat{j} + b\hat{k})$ and $\overrightarrow{r} = (\hat{i} - \hat{j} - 6\hat{k}) + \mu(-b\hat{i} + a\hat{j} + 5\hat{k})$ is $\frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4}$,then $a+b+c+d$ is equal to :