If the system of equations $x + 5y + 6z = 4$, $2x + 3y + 4z = 7$, and $x + 6y + az = b$ has infinitely many solutions, then the point $(a, b)$ lies on the line:

The system $x+2y+3z=4$,$4x+5y+3z=5$,$3x+4y+3z=\lambda$ is consistent and $3\lambda=n+100$,then $n=$

Let for any three distinct consecutive terms $a, b, c$ of an $A.P.$,the lines $ax + by + c = 0$ be concurrent at the point $P$ and $Q(\alpha, \beta)$ be a point such that the system of equations $x + y + z = 6$,$2x + 5y + \alpha z = \beta$ and $x + 2y + 3z = 4$ has infinitely many solutions. Then $(PQ)^2$ is equal to . . . . . . .

Consider the following system of equations: $\alpha x + 2y + z = 1$; $2\alpha x + 3y + z = 1$; $3x + \alpha y + 2z = \beta$. For some $\alpha, \beta \in \mathbb{R}$. Which of the following is $NOT$ correct?

If $A$ is a matrix such that $\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] A \left[\begin{array}{ll} 1 & 1 \end{array}\right] = \left[\begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array}\right]$, then $A$ is equal to

If the system of linear equations $x - 2y + kz = 1$,$2x + y + z = 2$,and $3x - y - kz = 3$ has a non-zero solution $(x, y, z) \neq 0$,then $(x, y)$ lies on the straight line whose equation is