If $P$, $Q$ and $R$ are three singular matrices given by $P = \begin{bmatrix} 2 & 3a \\ 4 & 3 \end{bmatrix}$, $Q = \begin{bmatrix} b & 5 \\ 2a & 6 \end{bmatrix}$ and $R = \begin{bmatrix} a^2 + b^2 - c & 1 - c \\ c + 1 & c \end{bmatrix}$, then the value of $(2a + 6b + 17c)$ is

A singular matrix is a matrix whose determinant equals zero. We'll find the determinant of each matrix and set it equal to 0.

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For matrix $P$, the determinant is:

$\text{det}(P) = (2)(3) - (3a)(4) = 0$

$6 - 12a = 0$

$a = \dfrac{1}{2}$

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For matrix $Q$, the determinant is:

$\text{det}(Q) = (b)(6) - (5)(2a) = 0$

$6b - 10a = 0$

Substitute $a = \dfrac{1}{2}$:

$6b - 10\left(\dfrac{1}{2}\right) = 0$

$6b - 5 = 0$

$b = \dfrac{5}{6}$

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For matrix $R$, the determinant is:

$\text{det}(R) = (a^2 + b^2 - c)(c) - (1 - c)(c + 1) = 0$

Expanding:

$a^2c + b^2c - c^2 - (c + 1 - c^2 - c) = 0$

$a^2c + b^2c - c^2 - 1 + c^2 = 0$

$a^2c + b^2c - 1 = 0$

$c = \dfrac{1}{a^2 + b^2}$

Substitute $a = \dfrac{1}{2}$ and $b = \dfrac{5}{6}$:

$a^2 = \dfrac{1}{4}$, $b^2 = \dfrac{25}{36}$

$a^2 + b^2 = \dfrac{9}{36} + \dfrac{25}{36} = \dfrac{34}{36} = \dfrac{17}{18}$

$c = \dfrac{1}{\dfrac{17}{18}} = \dfrac{18}{17}$

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Calculating $(2a + 6b + 17c)$:

$2a + 6b + 17c = 2\left(\dfrac{1}{2}\right) + 6\left(\dfrac{5}{6}\right) + 17\left(\dfrac{18}{17}\right)$

$= 1 + 5 + 18$

$= 24$