Let $P(x_1, y_1, z_1)$ be the foot of the perpendicular drawn from the point $Q(2, -2, 1)$ to the plane $x - 2y + z = 1$. If $d$ is the perpendicular distance from the point $Q$ to the plane and $l = x_1 + y_1 + z_1$,then the value of $l + 3d^2$ is:

Let the plane $P$ contain the line $2x+y-z-3=0=5x-3y+4z+9$ and be parallel to the line $\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$. Then the distance of the point $A(8,-1,-19)$ from the plane $P$ measured parallel to the line $\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}$ is equal to $............$.

If the line $\bar{r}=(\hat{i}-2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+\hat{j}+2 \hat{k})$ is parallel to the plane $\bar{r} \cdot(3 \hat{i}-2 \hat{j}-m \hat{k})=5$,then the value of $m$ is:

If the point of intersection of the lines $\frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7}$ and $\frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7}$ lies on the $xy$-plane, then the value of $a+b$ is:

The coordinates of the point,where the line through $A(3, 4, 1)$ and $B(5, 1, 6)$ crosses the $XZ$-plane,are

Find the vector equation of the line passing through $(1, 2, 3)$ and parallel to the planes $\vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = 5$ and $\vec{r} \cdot (3\hat{i} + \hat{j} + \hat{k}) = 6$.