If the vectors $\overrightarrow{a}=\hat{i}+a \hat{j}+a^{2} \hat{k}$,$\overrightarrow{b}=\hat{i}+b \hat{j}+b^{2} \hat{k}$ and $\overrightarrow{c}=\hat{i}+c \hat{j}+c^{2} \hat{k}$ are three non-coplanar vectors and $\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \\ b & b^{2} & 1+b^{3} \\ c & c^{2} & 1+c^{3}\end{array}\right|=0$,then the value of $abc$ is

If $a$ is perpendicular to $b$ and $c$,$|a| = 2$,$|b| = 3$,$|c| = 4$ and the angle between $b$ and $c$ is $\frac{2\pi}{3}$,then $[a \; b \; c]$ is equal to (in $\sqrt{3}$):

If the points $A(3,2,1)$,$B(4, x, 5)$,$C(4,2,-2)$,and $D(6,5,-1)$ are coplanar,then $x$ has the value:

If $a = i - j + k$,$b = i + 2j - k$ and $c = 3i + pj + 5k$ are coplanar,then the value of $p$ is:

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector,assuming the question implies the vectors are linearly dependent or part of a coplanar set,find $\lambda$. Given the standard form of such problems,if we consider the vectors $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ to be coplanar with a reference vector,let us assume the third vector is $\hat{k}$. For these to be coplanar,the scalar triple product must be zero: $\left|\begin{array}{ccc} 1 & -3 & 1 \\ \lambda & 3 & 0 \\ 0 & 0 & 1 \end{array}\right| = 0$. Solving this,$\lambda$ is equal to:

Observe the following statements:
$A$. Three vectors are coplanar if one of them is expressible as a linear combination of the other two.
$R$. Any three coplanar vectors are linearly dependent.
Then, which of the following is true?