All the real values of $p, q$ so that the system of equations $\begin{cases} 2x + py + 6z = 8 \\ x + 2y + qz = 5 \\ x + y + 3z = 4 \end{cases}$ may have no solution are

Statement $-1$: The system of linear equations
$x + (\sin \alpha)y + (\cos \alpha)z = 0$
$x + (\cos \alpha)y + (\sin \alpha)z = 0$
$x - (\sin \alpha)y - (\cos \alpha)z = 0$
has a non-trivial solution for only one value of $\alpha$ lying in the interval $(0, \frac{\pi}{2})$.
Statement $-2$: The equation in $\alpha$
$\left| \begin{matrix} \cos \alpha & \sin \alpha & \cos \alpha \\ \sin \alpha & \cos \alpha & \sin \alpha \\ \cos \alpha & -\sin \alpha & -\cos \alpha \end{matrix} \right| = 0$
has only one solution lying in the interval $(0, \frac{\pi}{2})$.

The cost of $4 \, kg$ onion,$3 \, kg$ wheat and $2 \, kg$ rice is $Rs \, 60$. The cost of $2 \, kg$ onion,$4 \, kg$ wheat and $6 \, kg$ rice is $Rs \, 90$. The cost of $6 \, kg$ onion,$2 \, kg$ wheat and $3 \, kg$ rice is $Rs \, 70$. Find the cost of each item per $kg$ using the matrix method.

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

If the system of linear equations $2x + y - z = 7$,$x - 3y + 2z = 1$,and $x + 4y + \delta z = k$,where $\delta, k \in R$,has infinitely many solutions,then $\delta + k$ is equal to

Solve the system of linear equations using the matrix method: $2x + 3y + 3z = 5$,$x - 2y + z = -4$,$3x - y - 2z = 3$.