The value of $k$,such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies on the plane $2x-4y+z=7$,is

If the equation of the plane passing through the point $(2,-3,4)$ and perpendicular to both the planes $2x-3y+5z=2$ and $x+y+2z=3$ is $x+py+qz=r$,then $r$ is equal to

Let a line with direction ratios $a, -4a, -7$ be perpendicular to the lines with direction ratios $3, -1, 2b$ and $b, a, -2$. If the point of intersection of the line $\frac{x+1}{a^{2}+b^{2}}=\frac{y-2}{a^{2}-b^{2}}=\frac{z}{1}$ and the plane $x - y + z = 0$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to $.......$

If the equation of a plane $P,$ passing through the intersection of the planes $x+4y-z+7=0$ and $3x+y+5z=8$ is $ax+by+6z=15$ for some $a, b \in R,$ then the distance of the point $(3,2,-1)$ from the plane $P$ is

If $\vec{a}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=-\hat{k}$ are position vectors of two points,and $\vec{b}=2 \hat{i}-\hat{j}+\lambda \hat{k}$ and $\vec{d}=\hat{i}+2 \hat{j}-\hat{k}$ are two vectors,then the lines $\vec{r}=\vec{a}+t \vec{b}$ and $\vec{r}=\vec{c}+s \vec{d}$ are:

Let $P_1: 2x + y - z = 3$ and $P_2: x + 2y + z = 2$ be two planes. Then,which of the following statement$(s)$ is (are) $TRUE$?
$(A)$ The line of intersection of $P_1$ and $P_2$ has direction ratios $1, -1, 1$.
$(B)$ The line $\frac{3x - 4}{9} = \frac{1 - 3y}{9} = \frac{z}{3}$ is perpendicular to the line of intersection of $P_1$ and $P_2$.
$(C)$ The acute angle between $P_1$ and $P_2$ is $60^{\circ}$.
$(D)$ If $P_3$ is the plane passing through the point $(4, 2, -2)$ and perpendicular to the line of intersection of $P_1$ and $P_2$,then the distance of the point $(2, 1, 1)$ from the plane $P_3$ is $\frac{2}{\sqrt{3}}$.