answer:We can write the equation as [frac{sin A}{cos A} + frac{1}{cos A} = 2,]so sin A + 1 = 2 cos A. Then sin A = 2 cos A - 1. Squaring both sides, we get [sin^2 A = 4 cos^2 A - 4 cos A + 1.]Since cos^2 A + sin^2 A = 1, [1 - cos^2 A = 4 cos^2 A - 4 cos A + 1,]which simplifies to 5 cos^2 A - 4 cos A = cos A (5 cos A - 4) = 0. Hence, cos A = 0 or cos A = frac{4}{5}. If cos A = 0, then sec A = frac{1}{cos A} is not defined. On the other hand, if A is the acute angle such that cos A = frac{4}{5}, then sin A = frac{3}{5}, so [tan A + sec A = frac{sin A + 1}{cos A} = frac{3/5 + 1}{4/5} = 2.]Therefore, cos A = boxed{frac{4}{5}}.

answer:We can write the equation of the second plane as x - 3y + 3z = 1. Note that (1,0,0) is a point on this plane. (Also, note that both plane have the same normal vector, so they are parallel.) Therefore, from the formula for the distance between a point and a plane, the distance between the two planes is [frac{|1 - 3 cdot 0 + 3 cdot 0 - 8|}{sqrt{1^2 + (-3)^2 + 3^2}} = boxed{frac{7 sqrt{19}}{19}}.]

answer:Let [f(x,y,z) = |x + y + z| + |x + y - z| + |x - y + z| + |-x + y + z|.]Note that begin{align*} f(-x,y,z) &= |-x + y + z| + |-x + y - z| + |-x - y + z| + |x + y + z| &= |-x + y + z| + |x - y + z| + |x + y - z| + |x + y + z| &= f(x,y,z). end{align*}Similarly, we can prove that f(x,-y,z) = f(x,y,-z) = f(x,y,z). This says that the set of points that satisfy [f(x,y,z) le 4]is symmetric with respect to the xy-, xz-, and yz-planes. So, we restrict our attention to the octant where all the coordinates are nonnegative. Suppose x ge y and x ge z. (In other words, x is the largest of x, y, and z.) Then begin{align*} f(x,y,z) &= |x + y + z| + |x + y - z| + |x - y + z| + |-x + y + z| &= 3x + y + z + |-x + y + z|. end{align*}By the Triangle Inequality, |-x + y + z| = |x - (y + z)| ge x - (y + z), so [f(x,y,z) = 3x + y + z + |-x + y + z| ge 3x + y + z + x - (y + z) = 4x.]But f(x,y,z) le 4, so x le 1. This implies that each of x, y, z is at most 1. Also, |-x + y + z| ge (y + z) - x, so [f(x,y,z) = 3x + y + z + |-x + y + z| ge 3x + y + z + (y + z) - x = 2x + 2y + 2z.]Hence, x + y + z le 2. Conversely, if x le 1, y le 1, z le 1, and x + y + z le 2, then [f(x,y,z) le 4.]The region defined by 0 le x, y, z le 1 is a cube. The equation x + y + z = 2 corresponds to the plane which passes through (0,1,1), (1,0,1), and (1,1,0), so we must cut off the pyramid whose vertices are (0,1,1), (1,0,1), (1,1,0), and (1,1,1). [asy] import three; size(180); currentprojection = perspective(6,3,2); draw(surface((0,1,1)--(1,0,1)--(1,1,0)--cycle),gray(0.8),nolight); draw(surface((1,0,0)--(1,1,0)--(1,0,1)--cycle),gray(0.6),nolight); draw(surface((0,1,0)--(1,1,0)--(0,1,1)--cycle),gray(0.7),nolight); draw(surface((0,0,1)--(1,0,1)--(0,1,1)--cycle),gray(0.9),nolight); draw((1,0,0)--(1,1,0)--(0,1,0)--(0,1,1)--(0,0,1)--(1,0,1)--cycle); draw((0,1,1)--(1,0,1)--(1,1,0)--cycle); draw((0,1,1)--(1,1,1),dashed); draw((1,0,1)--(1,1,1),dashed); draw((1,1,0)--(1,1,1),dashed); draw((0,0,0)--(1,0,0),dashed); draw((0,0,0)--(0,1,0),dashed); draw((0,0,0)--(0,0,1),dashed); draw((1,0,0)--(1.2,0,0),Arrow3(6)); draw((0,1,0)--(0,1.2,0),Arrow3(6)); draw((0,0,1)--(0,0,1.2),Arrow3(6)); label("x", (1.3,0,0)); label("y", (0,1.3,0)); label("z", (0,0,1.3)); [/asy] This pyramid has volume frac{1}{3} cdot frac{1}{2} cdot 1 = frac{1}{6}, so the remaining volume is 1 - frac{1}{6} = frac{5}{6}. Since we are only looking at one octant, the total volume of the region is 8 cdot frac{5}{6} = boxed{frac{20}{3}}.

answer:One way to solve for p, q, and r is to write p mathbf{a} + q mathbf{b} + r mathbf{c} as a three-dimensional vector, set the components to begin{pmatrix} -4 7 3 end{pmatrix}, and then solve the linear system. But we can also take advantage of the fact that mathbf{a} = begin{pmatrix} 1 1 1 end{pmatrix}, mathbf{b} = begin{pmatrix} 2 -3 1 end{pmatrix}, and mathbf{c} = begin{pmatrix} 4 1 -5 end{pmatrix} are mutually orthogonal. We can take the equation, begin{pmatrix} -4 7 3 end{pmatrix} = p mathbf{a} + q mathbf{b} + r mathbf{c}, and take the dot product of mathbf{a} with both sides: [mathbf{a} cdot begin{pmatrix} -4 7 3 end{pmatrix} = p mathbf{a} cdot mathbf{a} + q mathbf{a} cdot mathbf{b} + r mathbf{a} cdot mathbf{c}.]Note that mathbf{a} cdot mathbf{b} = mathbf{a} cdot mathbf{c} = 0, and we are left with [6 = 3a.]Hence, a = 2. In the same way, we can find b = -frac{13}{7} and c = -frac{4}{7}, so (a,b,c) = boxed{left( 2, -frac{13}{7}, -frac{4}{7} right)}.