The equation of the plane passing through the origin and perpendicular to the line $x = 2y = 3z$ is

Let a line $l$ pass through the origin and be perpendicular to the lines $l_1: \overrightarrow{r} = (\hat{i} - 11\hat{j} - 7\hat{k}) + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$ and $l_2: \overrightarrow{r} = (-\hat{i} + \hat{k}) + \mu(2\hat{i} + 2\hat{j} + \hat{k})$. If $P$ is the point of intersection of $l$ and $l_1$,and $Q(\alpha, \beta, \gamma)$ is the foot of the perpendicular from $P$ on $l_2$,then $9(\alpha + \beta + \gamma)$ is equal to:

Let $P$ be the image of the point $(3,1,7)$ with respect to the plane $x-y+z=3$. Then the equation of the plane passing through $P$ and containing the straight line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$ is

The line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta=0$. Then the value of $\alpha \beta$ is:

Let the lines $l_1: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}$ and $l_2: 3x+2y+z-2=0=x-3y+2z-13$ be coplanar. If the point $P(a, b, c)$ on $l_1$ is nearest to the point $Q(-4, -3, 2)$,then $|a|+|b|+|c|$ is equal to

If the lines $x = 1 + s, y = -3 - \lambda s, z = 1 + \lambda s, s \in R$ and $x = \frac{t}{2}, y = 1 + t, z = 2 - t, t \in R$ are coplanar,then $\lambda = $