KCET 2018

For the probability distribution given by        
$$ \begin{array}{|c|c|c|c|} \hline X=x_i & 0 & 1 & 2 \\ \hline P_i & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\ \hline \end{array} $$
the standard deviation $(\sigma)$ is

The equation of the line parallel to the line $3 x-4 y+2=0$ and passing through $(-2,3)$ is

The image of the point $(1,6,3)$ in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ is

The angle between the lines $2 x=3 y=-z$ and $6 x=-y=-4 z$ is

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

The locus represented by $x y+y z=0$ is

If the vector $a \hat{i}+\hat{j}+\hat{k} ; \hat{i}+b \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+c \hat{k}$ are coplanar $(a \neq b \neq c \neq 1)$, then the value of $a b c-(a+b+c)$ is equal to

If $|\vec{a} \times \vec{b}|^2+|\vec{a} \cdot \vec{b}|^2=144$ and $|\vec{a}|=4$, then the value of $|\vec{b}|$ is

If $\vec{a}$ and $\vec{b}$ are mutually perpendicular unit vectors, then $(3 \vec{a}+2 \vec{b}) \cdot(5 \vec{a}-6 \vec{b})$ is equal to

If $\vec{a}=\hat{i}+\lambda \hat{j}+2 \hat{k} ; \vec{b}=\mu \hat{i}+\hat{j}-\hat{k}$ are orthogonal and $|\vec{a}|=|\vec{b}|$, then $(\lambda, \mu)$ is equal to