The system of linear equations $x + y + z = 6$,$2x + 5y + az = 36$,and $x + 2y + 3z = b$ has:

  • A
    unique solution for $a = 8$ and $b = 16$
  • B
    infinitely many solutions for $a = 8$ and $b = 14$
  • C
    infinitely many solutions for $a = 8$ and $b = 16$
  • D
    unique solution for $a = 8$ and $b = 14$

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The linear system of equations $\begin{cases} 8x - 3y - 5z = 0 \\ 5x - 8y + 3z = 0 \\ 3x + 5y - 8z = 0 \end{cases}$ has

Let $\alpha, \beta$ and $\gamma$ be real numbers such that the system of linear equations
$x+2y+3z=\alpha$
$4x+5y+6z=\beta$
$7x+8y+9z=\gamma$
is consistent. Let $|M|$ represent the determinant of the matrix
$M=\begin{bmatrix} \alpha & 2 & \gamma \\ \beta & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$
Let $P$ be the plane containing all those $(\alpha, \beta, \gamma)$ for which the above system of linear equations is consistent,and $D$ be the square of the distance of the point $(0,1,0)$ from the plane $P$.
$(1)$ The value of $|M|$ is
$(2)$ The value of $D$ is

Investigate the values of $\lambda$ and $\mu$ for the system $x+2y+3z=6, x+3y+5z=9, 2x+5y+\lambda z=\mu$ and match the values in List-$I$ with the items in List-$II$.
List-$I$List-$II$
$(A)$ $\lambda=8, \mu \neq 15$$1$. Infinitely many solutions
$(B)$ $\lambda \neq 8, \mu \in R$$2$. No solution
$(C)$ $\lambda=8, \mu=15$$3$. Unique solution

For which of the following ordered pairs $(\mu, \delta)$ is the system of linear equations $x+2y+3z=1$,$3x+4y+5z=\mu$,and $4x+4y+4z=\delta$ inconsistent?

Let $\alpha, \beta$ and $\gamma$ be real numbers. Consider the following system of linear equations:
$x+2y+z=7$
$x+\alpha z=11$
$2x-3y+\beta z=\gamma$
Match each entry in List-$I$ to the correct entries in List-$II$:
List-$I$ List-$II$
$(P)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma=28$,then the system has $(1)$ a unique solution
$(Q)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma \neq 28$,then the system has $(2)$ no solution
$(R)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$,then the system has $(3)$ infinitely many solutions
$(S)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma=28$,then the system has $(4)$ $x=11, y=-2$ and $z=0$ as a solution
$(5)$ $x=-15, y=4$ and $z=0$ as a solution

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